Ch 16. Balanced Search Trees
© copyright 2006 SNU IDB Lab.
SNU
IDB Lab.
Bird’s-Eye View (0)

Chapter 15: Binary Search Tree


Chapter 16: Balanced Search Tree



BST and Indexed BST
AVL tree: BST + Balance
B-tree: generalized AVL tree
Chapter 17: Graph
Data Structures
2
SNU
IDB Lab.
Bird’s-Eye View

Balanced tree structures
- Height is O(log n)

AVL

Binary Search Tree with Balance

Red-black trees

Splay trees



Individual dictionary operation  0(n)
Take less time to perform a sequence of u operations  0(u log u)
B-trees (Balanced Tree)

Suitable for external memory
Data Structures
3
SNU
IDB Lab.
Table of Contents

AVL TREES




Definition
Searching an AVL Search Tree
Inserting into an AVL Search Tree
Deletion from an AVL Search Tree

RED-BLACK TREES

SPLAY TREES

B-TREES
Data Structures
4
SNU
IDB Lab.
The History of Balanced Trees

Adel'son-Vel'skiĭ and Landis introduced AVL tree in 1962



Bayer and McCreight introduced B-tree in 1972

Kept balanced by requiring that all leaf nodes are at the same depth

Join or split is needed instead of re-balancing
Bayer, Guibas and Sedgewick introduced Red-black tree in 1978


Ensures balance by restricting every node's depth to differ at most by 1
Ensures balance by restricting the occurrence of red nodes in the tree
Sleator and Tarjan introduced Splay tree in 1983

Maintains balance without any explicit balance condition such as color

Splay operations are performed within the tree every time an access is made
Data Structures
5
SNU
IDB Lab.
AVL TREES

Balanced tree


AVL search tree



Balanced binary search trees
Can be generalized to a B-tree
A height-balanced k tree (HB(k) tree)


Trees with a worst-case height of O(log n)
Allowable height difference of any two sub-trees is k
AVL Tree : HB(1) Tree

G.M. Adel’son, Vel’skii, E.M. Landis

Performance
Given N keys, worst-case search  1.44 log2(N+2)
cf. Completely balanced AVL tree : worst-case search  log2(N+1)

Data Structures
6
SNU
IDB Lab.
Height of an AVL Tree



n : nodes in AVL tree
Nh : min number of nodes in an AVL tree of height h
Nh = Nh-1 + Nh-2 + 1, N0 = 0, and N1 = 1
 Similar in definition to Fibonacci numbers
Fh = Fn-1 + Fn-2., F0 = 0 and F1 = 1


It can be shown that Nh = Fh+2 - 1 for h > 0
Fibonacci theory: Fh ≒ Øh/√5 where Ø = (1 + √5)/2
therefore Nh ≒ Øh+2/√5-1
If there are n nodes then its height h = logØ(√5(n+1)) - 2 ≒
1.44log2(n+2)
h = O(log n)
Data Structures
7
SNU
IDB Lab.
AVL Tree Definition


An empty binary tree is an AVL Tree
If T is a nonempty binary tree with TL and TR as its left and right subtrees,
then T is an AVL tree iff
(1) TL and TR are AVL Trees and
(2) | hL - hR| ≤ 1 where hL and hR are the heights of TL and TR, respectively

For any node in tree T in AVL tree, BF(T) should be one of “ -1, 0, 1”

If BF(T) is -2 or 2, then proper rotation is performed in order to get balance

Conceptually AVL search tree = AVL tree + Binary Search Tree
Data Structures
8
SNU
IDB Lab.
AVL Tree Examples
(a) AVL Trees
(b) Non - AVL Trees
X
X
X
Data Structures
9
X
SNU
IDB Lab.
Intuition: AVL Search Tree

AVL Search Tree = Binary Search Tree + AVL Tree
= Balanced Binary Search Tree
20
30
60
15
12
5
25
22
18
40
2
70
65
80
(a)
(b)
(c)
BST
X
O
O
AVL
O
O
X
AVL ST
X
O
X
Data Structures
10
SNU
IDB Lab.
Indexed AVL Search Tree

Indexed AVL search Tree
= AVL Tree + LeftSize variable
= (Balanced + Binary Search Tree) + LeftSize variable
3
1
1
Data Structures
APR
MAY
AUG
0
1
MAR
NOV
11
SNU
IDB Lab.
Representation of an AVL Tree


Balance factor bf(x) of a node x = height of left subtree – height of right subtree
Permissible balance factors: (-1, 0, 1)
20 0
30 -1
5 0
40
35 0
Data Structures
25 -1
0 15
1
0 12
12
18 0
30 0
SNU
IDB Lab.
AVL Search Tree Example (1)
New Identifier
After Insertion
MARCH
0
MAR
New Identifier
After Insertion
MAY
No Rebalancing needed
No Rebalancing needed
-1
MAR
0
MAY
New Identifier
NOVEMBER
After Rebalancing
After Insertion
-2
MAR
-1
MAY
Data Structures
0
MAY
RR
0
MAR
0
NOV
13
0
NOV
SNU
IDB Lab.
AVL Search Tree Example (2)
New Identifier
After Insertion
AUGUST
+1
MAY
+1
MAR
No Rebalancing needed
0
NOV
0
AUG
Data Structures
14
SNU
IDB Lab.
AVL Search Tree Example (3)
New Identifier
After Insertion
APRIL
+2
MAY
+2
MAR
After Rebalancing
LL
+1
MAY
0
NOV
0
AUG
+1
AUG
0
APR
0
NOV
0
MAR
0
APR
Data Structures
15
SNU
IDB Lab.
AVL Search Tree Example (4)
New Identifier
After Rebalancing
After Insertion
+2
MAY
JANUARY
-1
AUG
0
APR
0
MAR
LR
0
NOV
0
AUG
+1
MAR
0
APR
-1
MAY
0
JAN
0
NOV
0
JAN
Data Structures
16
SNU
IDB Lab.
AVL Search Tree Example (5)
New Identifier
No Rebalancing needed
After Insertion
+1
MAR
DECEMBER
-1
AUG
-1
MAY
+1
JAN
0
APR
0
NOV
0
DEC
Data Structures
17
SNU
IDB Lab.
AVL Search Tree Example (6)
New Identifier
+1
MAR
JULY
-1
AUG
-1
MAY
0
JAN
0
APR
0
DEC
Data Structures
No Rebalancing needed
After Insertion
0
NOV
0
JUL
18
SNU
IDB Lab.
AVL Search Tree Example (7)
New Identifier
FEBRUARY
RL
+2
MAR
-2
AUG
+1
MAR
-1
MAY
+1
JAN
0
APR
After Rebalancing
After Insertion
0
DEC
0
NOV
0
JAN
+1
AUG
0
JUL
-1
DEC
0
APR
-1
MAY
0
FEB
0
NOV
0
JUL
0
FEB
Data Structures
19
SNU
IDB Lab.
AVL Search Tree Example (8)
JUNE
0
APR
LR
+2
MAR
-1
DEC
0
FEB
0
JAN
+1
DEC
-1
MAY
-1
JAN
+1
AUG
After Rebalancing
After Insertion
New Identifier
+1
AUG
0
NOV
0
APR
-1
JUL
0
FEB
0
MAR
-1
MAY
-1
JUL
0
JUN
-1
NOV
0
JUN
Data Structures
20
SNU
IDB Lab.
AVL Search Tree Example (9)
After Rebalancing
After Insertion
New Identifier
0
JAN
OCTOBER-1
JAN
+1
DEC
+1
AUG
RR
-1
MAR
0
FEB
0
APR
-1
JUL
0
JUN
0
MAR
+1
DEC
+1
AUG
-2
MAY
-1
NOV
0
APR
0
FEB
-1
JUL
0
JUN
0
NOV
0
MAY
0
OCT
0
OCT
Data Structures
21
SNU
IDB Lab.
AVL Search Tree Example (11)
New Identifier
No Rebalancing needed
After Insertion
SEPTEMBER
-1
JAN
-1
MAR
+1
DEC
+1
AUG
0
APR
0
FEB
-1
JUL
0
JUN
-1
NOV
0
MAY
-1
OCT
0
SEP
Data Structures
22
SNU
IDB Lab.
Table of Contents

AVL TREES

Definition

Searching an AVL Search Tree


Inserting into an AVL Search Tree
Deletion from an AVL Search Tree

RED-BLACK TREES

SPLAY TREES

B-TREES
Data Structures
23
SNU
IDB Lab.
Searching in an AVL Search Tree

search in binary search tree : Wish to Search for thekey from root to leaf
If (root == null) search is unsuccessful;
else
if
(thekey < key in root) only left subtree is to be searched;
else if (thekey > key in root) only right subtree is to be searched;
else (thekey == key in root) search terminates successfully;

Subtrees may be searched similarly in a recursive manner

TimeComplexity = O(height)

Height of an AVL tree with n element

O(log n): search time is O(log n)
Data Structures
24
SNU
IDB Lab.
Table of Contents

AVL TREES




Definition
Searching an AVL Search Tree
Inserting into an AVL Search Tree
Deletion from an AVL Search Tree

RED-BLACK TREES

SPLAY TREES

B-TREES
Data Structures
25
SNU
IDB Lab.
Unbalance due to Inserting

When an insertion into an AVL Tree using the strategy of Program15.5
(insert in BST), the resulting tree is unbalanced
30
5 0
0
-1
40
35
Data Structures
1
New
element
26
SNU
IDB Lab.
Observations on
Imbalance due to Insertion
O1: In the unbalanced tree the BFs are limited to –2, -1, 0, 1, 2
O2: A node with BF “2” had a BF “1” before the insertion
O3: The BF of only those nodes on the path from the root to the newly
inserted node can change as a result of the insertion
O4: Let A denote the nearest ancestor of the newly inserted node whose
BF is either –2 or 2. The BF of all nodes on the path from A to the
newly inserted node was 0 prior to the insertion
O5: Imbalance can happen in the last node encountered that has a balance factor
1 or –1 prior to the insertion
Data Structures
27
SNU
IDB Lab.
Node X with Potential Imbalance (1)


Let X denote the last node encountered that has a balance factor 1 or –1
prior to the insertion
If the tree is unbalanced following the insertion, X exists

If bf(x) = 0 after the insertion, then the height of the subtree with root X is
the same before and after the insertion
No node X
-1 30
0 5
0 20
X
40 1
35 0
32
Data Structures
0 15
12
0
0 20
X
-125
18 22
0 0 30
28 50
28
0 15
0 12
-125
18 0 0 30
10 14 16 19
SNU
IDB Lab.
Node X with Potential Imbalance (2)
(a)
(b)
(c)
height
h
h
h+1
bf(x)
1
0
2
balanced
balanced
balanced
imbalanced

The only way the tree can become unbalanced is when the insertion
causes bf(x) to change from –1 to –2 or from 1 to 2.
Data Structures
29
SNU
IDB Lab.
Imbalance Patterns due to Insertion

The imbalance at A is one of the types
LL (when new node is in the left subtree of the left subtree of A)

LR (when new node is in the right subtree of the left subtree of A)

RR (when new node is in the right subtree of the right subtree of A)

RL (when new node is in the left subtree of the right subtree of A)
LL and RR imbalances require single rotation
LR and RL imbalances require double rotations



A
Insert Y
LL
Data Structures
RL
LR
30
RR
SNU
IDB Lab.
LL Rebalancing after Insertion
Balanced Subtree
Unbalanced following
insertion
+1
A
+2
A
0
B
AR
0
B
h+2
rotation type
LL
BR
BL
0
B
BL
AR
0
A
h+2
h
BL
Balanced Subtree
BR
BR
AR
Height of BL increase to h+1
(BL < B < BR < A < AR)
Data Structures
31
SNU
IDB Lab.
RR Rebalancing after Insertion
Unbalanced following
insertion
Balanced Subtree
0
B
AL
BL
rotation type
RR
-2
A
-1
A
BR
0
A
h+2
h+2
BL
Al
BR
Height of BR increase to h+1
(AL < A < BL < B < BR)
Data Structures
0
B
0
B
AL
BR
Balanced Subtree
32
BL
SNU
IDB Lab.
LR-a Rebalancing after Insertion
Balanced Subtree
+1
A
0
B
Unbalanced following
insertion
rotation type
Balanced Subtree
0
C
LR(a)
+1
A
0
B
-1
B
0
A
0
C
(B < C < A)
Data Structures
33
SNU
IDB Lab.
LR-b Rebalancing after Insertion
Unbalanced following
insertion
Balanced Subtree
0
B
AR
h+2
BL
CR
h+2
-1
A
AR
+1
C
h
h-1 CL
0
C
0
B
-1
B
h
0
C
BL
rotation type
LR(b)
+2
A
+1
A
Balanced Subtree
CL
BL
CL
CR
AR
h
CR
(BL < B < CL < C < CR < A < AR)
Data Structures
34
SNU
IDB Lab.
LR-c Rebalancing after Insertion
Unbalanced following
insertion
Balanced Subtree
0
B
AR
h-1 CL
h+2
BL
CR
0
C
h+2
0
A
+1
B
-1
B
h
0
C
BL
rotation type
LR(c)
+2
A
+1
A
Balanced Subtree
AR
-1
C
CL
BL
CL
CR
AR
h
CR
RL a, b and c are symmetric to LR a, b and c
Data Structures
35
SNU
IDB Lab.
Table of Contents

AVL TREES




Definition
Searching an AVL Search Tree
Inserting into an AVL Search Tree
Deletion from an AVL Search Tree

RED-BLACK TREES

SPLAY TREES

B-TREES
Data Structures
36
SNU
IDB Lab.
Deletion from an AVL Tree

Let q be the parent of the node that was physically deleted

If the deletion took place from



the left subtree of q  bf(q) decreases by 1
the right subtree of q  bf(q) increases by 1
Observations
D1 : If the new BF of q is 0, its height has decreased by 1.
we need to change the BF of its parent (if any) and possibly those of its other
ancestors
D2 : If the new BF of q is either –1 or 1, its height is the same as before the
deletion and the BFs of tis ancestors and unchanged
D3 : If the new BF of q is either –2 or 2, the tree is unbalanced at q
Data Structures
37
SNU
IDB Lab.
Imbalance Patterns due to Deletion

Type L



If the deletion took place from A’s left subtree with root B
Subclassified : L-1, L0 and L1 depending on bf(B)
Type R


If the deletion took place from A’s right subtree with root B
Subclassified : R-1, R0 and R1 depending on bf(B)
Data Structures
38
SNU
IDB Lab.
R0 rotation after Deletion



Height of tree is h+2 (h+2) before (after) deletion
Single rotation is sufficient
BL < B < BR < A < AR
Data Structures
39
SNU
IDB Lab.
R1 rotation after Deletion



Height of tree is h+2 (h+1) before (after) deletion
Single rotation is sufficient
BL < B < BR < A < AR
Data Structures
40
SNU
IDB Lab.
R-1 rotation after Deletion



Height of tree is h+2 (h + 1) before (after) deletion
Double rotations
BL < B < CL < C < CR < A < AR
Data Structures
41
SNU
IDB Lab.
Rotation Taxonomy in AVL

Rotation types due to Insertion





Rotation types due to Deletion





LL type
RR type
LR type: LR-a, LR-b, LR-c
RL type: RL-a, RL-b, LR-c
R type: R-1, R0, R1
L type: L-1, L0, L1
LL rotation in insertion and R1 rotation in deletion are identical
LR rotation in insertion and R-1 rotation in deletion are identical
LL rotation in insertion and R0 rotation in deletion differ only in the final
BF of A and B
Data Structures
42
SNU
IDB Lab.
Table of Contents

AVL TREES

RED-BLACK TREES





Definition
Searching a Red-Black Tree
Inserting into a Red-Black Tree
Deletion from a Red-Black Tree
Implementation Considerations and Complexity

SPLAY TREES

B-TREES
Data Structures
43
SNU
IDB Lab.
Red-Black Tree vs. AVL Tree (1)
more balanced
less balanced
Data Structures
Red-Black tree
AVL tree
Lookup
O(logn)
O(logn)
Insertion
O(logn)
O(logn)
Deletion
O(logn)
O(logn)
44
SNU
IDB Lab.
Red-Black Tree vs. AVL Tree (2)
insert a node x
x
x
Red-black tree doesn't need rebalancing
Data Structures
AVL tree needs rebalancing
45
SNU
IDB Lab.
Red-Black Tree: Definition

Red-black tree
Binary Search tree
 Every node is colored red or black
RB1. Root and all external nodes are black.
RB2. No root-to-external-node path has two consecutive red nodes.
RB3. All root-to-external-node paths have the same number of black nodes

≡
equivalent
RB1’. Pointers from an internal node to an external node are black
RB2’. No root-to-external-node path has two consecutive red pointers
RB3’. All root-to-external-node paths have the same number of black
pointers
Data Structures
46
SNU
IDB Lab.
Red-Black Tree: Example
65
50
10
5




80
70
60
62
Every path from the root to an external node has exactly 2 black
pointers and 3 black nodes
No such path has two consecutive red nodes or pointers
Small black box nodes are for ensuring every node has two children
The color of newly inserted node is red
SNU
Data Structures
47
IDB Lab.
RBT: Glossary


Rank: number of black pointers on any path from the node to any
external node in red-black tree
Length (of a root-to-external-node path): number of pointers on the
path.
• rank = 1
• height = length = 2
Data Structures
48
SNU
IDB Lab.
RBT: Lemma 1

Lemma 1


If P and Q are two root-to-external-node
paths in a red-black tree,
Then length(P) ≤ 2 * length(Q)
Proof



Suppose that the rank of the root is r
From RB1’ and RB2’, each root-to-externalnode path has between r and 2r pointers
So length(P) ≤ 2length(Q)
Data Structures
49
length(Q)=2
length(P)=4
SNU
IDB Lab.
RBT: Lemma 2

Lemma 2
h=4
n=5
r=2
h : height of a red-black tree
n : number of internal nodes
r : rank of the root




(a) h ≤ 2r


(b) n ≥ 2r – 1


From Lemma 16.1, no root-to-external-node path has length > 2r
No external nodes at levels 1 through r so 2r – 1 internal nodes at these
levels
(c) h ≤ 2log2(n+1)



2r ≤ n + 1 from (b)
r ≤ log2(n+1)
f ≤ 2r ≤ 2log2(n+1)
Data Structures
50
SNU
IDB Lab.
RBT: Representation



Null pointers represent external nodes
Pointer and node colors are closely related
Each node we need to store only
its color ( one additional bit per node ) or
the color of the two pointers to its children
(two additional bit per node)
Data Structures
51
→ null pointer
→R/B
or
→ {R / B, R / B}
SNU
IDB Lab.
Table of Contents


AVL Tree
RED-BLACK TREES

Definition

Searching a Red-Black Tree





Inserting into a Red-Black Tree
Deletion from a Red-Black Tree
Implementation
Splay Tree
B Tree
Data Structures
52
SNU
IDB Lab.
Searching a Red-Black Tree

Use the same code to search ordinary binary search tree (Program 15.4),
AVL tree, red-black trees
if(root == null) {
search is unsuccessful
} else {
if ( thekey < key in root)
only left subtree is to be searched
} else {
if(thekey > key in root)
only right subtree is to be searched
else (thekey == key in root)
search terminates successfully
}
}
Data Structures
53
SNU
IDB Lab.
Table of Contents


AVL Tree
RED-BLACK TREES

Definition
Searching a Red-Black Tree

Inserting into a Red-Black Tree





Deletion from a Red-Black Tree
Implementation
Splay Tree
B Tree
Data Structures
54
SNU
IDB Lab.
Violations due to Insertion (1)


The RBT should have the same number of black nodes in all paths
If new node is colored as black

The updated tree will always violate RB3 (same number of black nodes)
3
3
2
insert 1
2
4
4
1
Data Structures
55
SNU
IDB Lab.
Violations due to Insertion (2)

If new node is colored as red


If the parent of inserted node is black, it's OK (no violation).
But if the parent of inserted node is also red, violation occurs!
Violate RB2 (no two consecutive reds)

3
3
insert 1
2
2
1
Data Structures
56
SNU
IDB Lab.
L Type Imbalances due to Insertion (1)

u be the inserted node (red)



pu be the parent of u (red)
 puL & puR
gu be the granparent of u


uL & uR
guL & guR
LLr & LRr

The color of guR is red
Data Structures
57
SNU
IDB Lab.
L Type Imbalances due to Insertion (2)

u be the inserted node (red)

uL & uR
U


pu be the parent of u (red)
 puL & puR
gu be the granparent of u


U
guL & guR
LLb & LRb

The color of guR is black
Data Structures
58
SNU
IDB Lab.
Fixing LLr and LRr Imbalance
Begin
change the color of pu & guR : red  black
if (gu != root) { change the color of gu : black  red }
else { the color change not done.
the number of black nodes increases by 1.
(on all root-to-external-node paths) }
if (the color change of gu causes imbalance)
gu became the new u node
if (gu != root && the color change causes imbalance)
continue to rebalance
End
Data Structures
59
SNU
IDB Lab.
Fixing LLr Imbalance
If a node (which is red) u is left child of its parent (also red)
and its parent is left child of its grandparent & its uncle is red,
then change its grandparent's color to red &
change its parent's and uncle's color to black
A
B
u C
D
B
G
F
E
LLr imbalance
Data Structures
A
u C
D
G
F
E
After LLr color change
60
SNU
IDB Lab.
Fixing LRr Imbalance
If a node (which is red) u is right child of its parent (also red)
and its parent is left child of its grandparent & its uncle is red,
then change its grandparent's color to red
& change its parent's and uncle's color to black
A
A
B
D u
C
B
G
E
LRr imbalance
Data Structures
F
C
G
D u
F
E
After LRr color change
61
SNU
IDB Lab.
Fixing LLb and LRb Imbalance

Rotation first & then Change the color




The root of the involved subtree is black following the rotation
Number of black nodes on all root-to-external-node paths is unchanged
LLb rotation in RB tree is similar to LL rotation in AVL tree
LRb rotation in RB tree is similar to LR rotation in AVL tree
Data Structures
62
SNU
IDB Lab.
Fixing LLb Imbalance
If a node (which is red) u is left child of its parent(also red)
and its parent is left child of its grandparent & its uncle is black,
then do rotation and color change like the following
B
A
u C
u
E
B
D
D
LLb imbalance
Data Structures
A
C
E
After LLb rotation
63
SNU
IDB Lab.
Fixing LRb Imbalance
a node (which is red) u is right child of its parent (also red)
and its parent is left child of its grandparent & its uncle is black,
then do rotation and color change like the following
If
A
B
u D
G
D u
C
E
C
F
LRb imbalance
Data Structures
A
B
E
F
G
After LRb rotation
64
SNU
IDB Lab.
Insertion Example in RBT (1)
(a) Initial state:
all root-to-external-node paths have
3 black nodes & 2 black pointers
50
80
10
90
50
80
10
70
Data Structures
(b) insert 70 as a red node:
No violations of RBT 
No remedial action is necessary
90
65
SNU
IDB Lab.
Insertion Example in RBT (2)
50
(c) insert 60 as a red node  LLr imbalance
gu
10
pu
uu
70
80
90
60
50
pu
80
10
70
(d) LLr color change on nodes 70, 80 & 90;
gu is null, so not RB2 imbalance
Data Structures
66
u
90
60
SNU
IDB Lab.
Insertion Example in RBT (3)
50
(e) Insertion 65 as a red node  LRb imbalance
80
10
gu
pu
70
90
60
u
50
65
80
10
65
(f) Perform LRb rotation
Data Structures
67
60
90
70
SNU
IDB Lab.
Insertion Example in RBT (4)
50
(g) Insertion 62 as a red node  LRr imbalance
80
10
gu
pu
65
60
u
90
gu 50
70
62
80
10
u
(h) LRr color change on nodes 65, 60 & 70 
RLb imbalance
Data Structures
68
65
60
pu
90
70
62
SNU
IDB Lab.
Insertion Example in RBT (5)
gu 50
80
10
u
65
60
pu
RLb imbalance
90
65
70
50
62
10
80
60
70
90
62
(i) Perform RLb rotation
Data Structures
69
SNU
IDB Lab.
Table of Contents


AVL Tree
RED-BLACK TREES

Definition
Searching a Red-Black Tree
Inserting into a Red-Black Tree

Deletion from a Red-Black Tree

Implementation




Splay Tree
B Tree
Data Structures
70
SNU
IDB Lab.
Violations due to Deletion (1)

If the parent of deleted node is red, RB2 violation occurs!
4
3
2
4
...
3
...
delete 2
1
1
Data Structures
71
SNU
IDB Lab.
Violations due to Deletion (2)

If the deleted node is black, RB3 violation occurs!
3
2
Data Structures
3
delete 2
4
4
72
SNU
IDB Lab.
Deletion & Imbalance in RBT (1)
65
50
10
(a) A Red-Black tree
90
60 70
62
65
50
10
60
62
Data Structures
90
(b) Delete 70

Deleted node was red

Same number of black nodes before and
after the rotation

This is OK
73
SNU
IDB Lab.
Deletion & Imbalance in RBT (2)
65
(a) A Red-Black tree
50
10
90
60 70
62
65
50
10
60
62
Data Structures
70
y
(c) Delete 90

The red node 70 takes the place of the deleted
node which was black

Then, the number of black nodes on path
from root-to-external node in y is 1 less than
before  RB3 violation occurs = imbalance

Change the color of y to Black
74
SNU
IDB Lab.
Deletion & Imbalance in RBT (3)
65
50
10
62
90
50
60 70
10
90
60 70
62
(a) A Red-Black tree
(d) Delete 65

Deleted node was black and the node 62
was red, so change to black
** An RB3 violation occurs
only when the deleted node was black
and y is not the root of the resulting tree.
Data Structures
75
SNU
IDB Lab.
Rb Imbalance due to Deletion



Rb0 => color change
Rb1 => handled by rotation
Rb2 => handled by rotation
number of y’s nephew
y's sibling is black
y is the right child of its parent
(y is the node that takes the place of removed node)
Data Structures
76
SNU
IDB Lab.
Deletion Imbalances:
Rb family
y: the node that takes the
place of removed node
py: parent of y
v: sibling of y
vL & vR: children of v
Data Structures
77
SNU
IDB Lab.
Fixing Rb0 Imbalance
If a node (which is black) y is right child of its parent
and its sibling is black & its sibling has 0 red child,
then change its sibling's color to red
A
A
E y
B
C
Data Structures
C
D
Rb0 imbalance
E y
B
D
After Rb0 color change
78
SNU
IDB Lab.
Fixing Rb1 Imbalance
If a node(which is black) y is right child of its parent
and its sibling is black & its sibling has 1 red child,
then do rotation and color change like the following
B
C
A
G y
D
F
E
A
red /
black
G y
B
D
B
D
C
E
C
F
Rb1 imbalance
Data Structures
A
E
F
G y
After Rb1 rotation
79
SNU
IDB Lab.
Fixing Rb2 Imbalance
If a node(which is black) y is right child of its parent
and its sibling is black & its sibling has 2 red children,
then do rotation and color change like the following
A
D
G y
B
B
D
C
E
F
Rb2 imbalance
Data Structures
C
A
E
F
G y
After Rb2 rotation
80
SNU
IDB Lab.
Rr Imbalance due to Deletion



Rr0
Rr1
Rr2
handled by rotation
number of red child that v’s right child has
(v is sibling of y)
y's sibling is red
y is the right child of its parent
(y is the node that takes the place of removed node)SNU
Data Structures
81
IDB Lab.
Deletion Imbalances:
Rr family
y: the node that takes the
place of removed node
py: parent of y
v: sibling of y
vL & vR: children of v
Data Structures
82
SNU
IDB Lab.
Fixing Rr0 Imbalance
If a node(which is black) y is right child of its parent
and its sibling is red & its nephew has 0 red child,
then do rotation and color change like the following
B
A
E y
B
C
D
Rr0 imbalance
Data Structures
A
C
D
E y
After Rr0 rotation
83
SNU
IDB Lab.
Fixing Rr1 Imbalance
If a node(which is black) y is right child of its parent
and its sibling is red & its nephew has 1 red child,
then do rotation and color change like the following
D
B
C
I y
red /
black
F
D
B
F
E
G
C
H
A
D
E
Rr1 imbalance
Data Structures
H
G
B
I y
F
E
A
C
A
H
I y
G
After Rr1 rotation
84
SNU
IDB Lab.
Fixing Rr2 Imbalance
If a node(which is black) y is right child of its parent
and its sibling is red & its nephew has 2 red children,
then do rotation and color change like the following
A
I y
B
F
B
D
C
C
F
E
G
Rr2 imbalance
Data Structures
D
E
H
A
H
I y
G
After Rr2 rotation
85
SNU
IDB Lab.
Deletion Example (1)
65
50
10
80
60
70
90
65
62
50
10



(a) 90 deleted
Not root & black
Imbalance Rb0
Data Structures
80
60
v
py
70
y
62
vR
86
SNU
IDB Lab.
Deletion Example (2)
65

80 py
50
10
60
v


70

(b) Rb0 color change
py was red before delete
Rb0 color change of 70 & 80
we are done
65
62
vR
50
10



( C) delete 80
Black node “80” was deleted
So tree remains balanced
Data Structures
70
60
62
87
SNU
IDB Lab.
Deletion Example (3)
py 65
65
50
10
v
70
10
60

62
50
60v


65


Data Structures
60v w
62 x
62
10
50
(d) delete 70
Nonroot black node was deleted
Tree is imbalance Rr1(ii)
(e) after Rr1(ii) Rotation
This tree is now balanced!
88
SNU
IDB Lab.
Rotation Taxonomy in RBT

Rotation types due to Insertion

L family


LLb type
LRb type
RRr type RLr type
RRb type
RLb type
R family


LLr type LRr type
Rotation types due to Deletion

Rb family


Rb1(ii)
Rb2
Rr0
Rr1(i)
Rr1(ii)
Rr2
Lb0
Lb1(i)
Lb1(ii)
Lb2
Lr0
Lr1(i)
Lr1(ii)
Lr2
Lb family


Rb1(i)
Rr family


Rb0
Lr family

Data Structures
89
SNU
IDB Lab.
Table of Contents


AVL Tree
RED-BLACK TREES

Definition
Searching a Red-Black Tree
Inserting into a Red-Black Tree
Deletion from a Red-Black Tree

Implementation





Splay Tree
B Tree
Data Structures
90
SNU
IDB Lab.
Implementation

Considerations



Insertion / Deletion require backward movement
If use red-black-tree nodes Backward movement is easy
else Backward movement is complex //use stack instance of color fields..etc
Complexity

For an n-element red-black tree





parent-pointer scheme runs slightly faster than tack scheme
Color change : O(log n) // propagate back toward the root
Rotation : O(1)
Each color change or ratation : Θ(1)
Total insert/delete O(log n)
Data Structures
91
SNU
IDB Lab.
Table of Contents

AVL TREES

RED-BLACK TREES

SPLAY TREES

B-TREES
Data Structures
92
SNU
IDB Lab.
Splay Tree


Splay tree is a binary search tree whose nodes are rearranged by splay
operation whenever search, insertion, or deletion occurs
The recently accessed node is moved to the top
Self-Balancing
by Splay operation

Properties of splay tree

Recently accessed elements are quick to access again



Basic operations run in O(log n) amortized time
It is simpler to implement splay trees than red-black trees or AVL trees
Splay trees don't need to store any extra data in nodes
SNU
Data Structures
93
IDB Lab.
The Splay Operation

We call recently accessed(searched, inserted, or deleted) node as splay node


Splay operation is performed on splay node to move it to the root
We can perform successive accesses faster because recently accessed node is moved to
the top of the tree
x
g
p
x
A
g
splay node
A
B

p
D
B
C
D
C
Splay operation comprises sequence of the following splay steps.
If (Splay node = root) then sequence of steps is empty
Else splay step moves the splay node either 1 level or 2 levels up the tree
Data Structures
94
SNU
IDB Lab.
Splay Node



Search(x) makes the node x as a splay node
Insert(x) makes the node x as a splay node
Delete(x) makes the parent node of x as a splay node
5
2
1
5
6
4
1
Search(4)
2
5
2
6
4
splay node
3
2
6
4
1
Insert(3)
5
splay node
6
1
splay node
Delete(4)
Data Structures
95
SNU
IDB Lab.
One Level Splay Step

When the level of splay node = 2 (Only)



L splay step : splay node is Left child of its parent
R splay step : splay node is Right child of its parent
L splay step


If splay node q is the left child of its parent, then do rotation like the
following
Notice that following the splay step the splay node becomes the root of
binary search tree
root
Data Structures
root
96
SNU
IDB Lab.
Two Level Splay Step


When the level of splay node > 2
Types




LL : p is Left child of gp, q is Left child of p
LR : p is Left child of gp, q is Right child of p
RR : p is Right child of gp, q is Right child of p
RL : p is Right child of gp, q is Left child of p
LL
Data Structures
RL
LR
97
RR
SNU
IDB Lab.
LL Splay Step


If splay node q is the left child of its parent
& its parent is the left child of its grandparent,
then do rotation like the following
The splay node is moved 2 level up
Data Structures
98
SNU
IDB Lab.
LR Splay Step


If splay node q is the right child of its parent
& its parent is the left child of its grandparent,
then do rotation like the following
The splay node is moved 2 level up
Data Structures
99
SNU
IDB Lab.
Sample Splay Operation
Search “2”
Data Structures
100
SNU
IDB Lab.
Rotation Taxonomy of Splay Tree

1 level splay step



L type
R type
2 level splay step




LL type
LR type
RL type
RR type
Data Structures
101
SNU
IDB Lab.
Concept of Amortized


Rule: Spend less than 100$ per month

Normal spending – Spend less than 100$ per month

Amortized spending – Spend less than (100 * 12)$ per year
Remember array expansion


Regular complexity

Double the size (initialize) -- O(n)

Copy the old array to the new array – O(n)
Amortized complexity

Doubling will happen after n insertions!

One insertion is responsible for one slot expansion  O(1)
Data Structures
102
SNU
IDB Lab.
Amortized Complexity (1)


In an amortized analysis, the time required to perform a sequence of
data-structure operations is averaged over all the operations
performed
Amortized analysis differs from average-case analysis

Amortized analysis guarantees the average performance of each
operation in the worst case


n
n
i 1
i 1
 amortized (i)   actual(i)
Theorem 16.1


The amortized complexity of a get, put or remove operation performed
on a splay tree with n element is O(log n)
Actual Complexity of any sequence of g get, p put and r remove
operations

O((g+p+r)log n)
Data Structures
103
SNU
IDB Lab.
Amortized Complexity (2)
Example (1)

7
6
5
4
4
2
1
4
1
2
splay
2
2
LL
1
5
4
3
6
5
splay
L
7
4
6
3
3
2
3
5
splay
LR
1
7
6
6
search(2)
5
1
7
7
3
T1 = (search time)+(splay time)= 6 comparisons + 5 rotations
Data Structures
104
SNU
IDB Lab.
Amortized Complexity (3)

Example (2)
2
1
7
1
search(1)
5
4
3
1
2
7
5
4
6
6
3
2
7
splay
L
5
4
6
3
T2 = (search time) + (splay time) = 2 comparisons + 1 rotation
Data Structures
105
SNU
IDB Lab.
Amortized Complexity (4)

Example (3)
1
1
2
2
7
7
5
4
3
1
2
search(2)
5
4
6
6
splay
R
7
5
4
6
3
3
T3 = (search time) + (splay time) = 2 comparisons + 1 rotations
Data Structures
106
SNU
IDB Lab.
Amortized Complexity (5)

Example (4)



In the previous example, total time taken is 10 comparisons + 7 rotations
If there were no splay operation, total time taken would be 18 comparisons
Generally, it is known that (t1+t2+…+tk) / k ≤ 3*log2n, where n is the
number of nodes if k is large enough
7
6
7
6
search(2)
5
7
5
4
4
1
2
3
5
4
1
2
6
search(2)
5
4
1
Data Structures
6
search(1)
7
1
2
3
2
3
107
3
SNU
IDB Lab.
Table of Contents

AVL TREES
RED-BLACK TREES
SPLAY TREES

B-TREES









Indexed Sequential Files (ISAM)
m-WAY Search Trees
B-Trees of Order m
Height of a B-Tree
Searching a B-Tree
Inserting into a B-Tree
Deletion from a B-Tree
Data Structures
108
SNU
IDB Lab.
Indexed Sequential Access Method (ISAM)


Small dictionary may reside in internal memory
Large dictionary must reside on a disk



ISAM file (= Indexed Sequential file)



A disk consists of many blocks
Elements (records) are packed into a block in ascending order
disk-based file structure for large dictionary
Provide good sequential and random access
Primary Concern: reducing the number of disk IO s during search
Data Structures
109
SNU
IDB Lab.
Overview : ISAM File
R
61 
b
101 
10 20 50 61
d
1 3 10
A B A
e
11 20
C D
f
30 40 45
D C A
part description records
primary key
PART No
PART-Type
a
g
51 55 57
A D B
50
D
c
h
65 70 101
E B C
60
B
i
120150
A D
61
A
Example : Indexed sequential structure (when using overflow chain)
File Structures
Data Structures
110
SNU
IDB Lab.
File Structure Evolution


Sequential file: records can be accessed sequentially
 not good for access, insert, delete records in random order
Indexed-sequential file = Indexed Sequential Access Method (ISAM)


Sequential file + Index
B+ tree file

Indexed-sequential file + Balance
But here we study “B tree” data structure
--- m-Way search tree is similar to ISAM file
File Structures
Data Structures
111
SNU
IDB Lab.
Table of Contents

AVL TREES
RED-BLACK TREES
SPLAY TREES

B-TREES









Indexed Sequential Files (ISAM)
m-WAY Search Trees
B-Trees of Order m
Height of a B-Tree
Searching a B-Tree
Inserting into a B-Tree
Deletion from a B-Tree
Data Structures
112
SNU
IDB Lab.
m-Way Search Tree




Binary Search Tree can be generalized to m-Way search tree
White box is an internal node while solid square is external node
Each internal node can have up to six keys and seven pointers
A certain input sequence would build the following example
Data Structures
113
SNU
IDB Lab.
Properties of m-WAY Search Tree
m-Way search tree has the following properties



In the corresponding extended search tree, each internal node has up to p+1 children
and between 1 and p elements.
Every node with p elements has exactly p + 1 children

Let k1, ...,kp be the keys of these ordered elements (k1< k2<…< kp)

Let c0, c1…, cp be the p+1 children of the node.
Key ranges




The elements in the subtree with root co have keys smaller than k1
The elements in the subtree with root cp have keys larger than kp
The elements in the subtree with root ci have keys larger than ki but smaller than ki+1,
1≤ i ≤ p
Data Structures
114
SNU
IDB Lab.
Searching an m-Way Search Tree

Search the element with key 31
 10< 31 <80 : Move to the middle subtree
 k2< 31 <k3 : Move to the third subtree
 31< k1
: Move to the first subtree, Fall off the tree, No element
Data Structures
115
SNU
IDB Lab.
Inserting into an m-Way Search Tree

Insert the new key 31
(a) Search for 31 & Fall off the tree at the node[32,26]
(b) Insert at the first element in the node
Data Structures
116
SNU
IDB Lab.
Inserting into an m-Way Search Tree

Insert the new key 65:
(a) Search for 65 & Fall off the tree at six subtree of node [20,30,40,50,60,70]
(b) New node obtained & New node becomes the sixth child of [20,30,40,50,60,70]
65
Data Structures
117
SNU
IDB Lab.
Deleting from an m-Way Search Tree

Delete the key 20

Search for 20, k1=20 & C0=C1=0, and Simply Delete 20
Data Structures
118
SNU
IDB Lab.
Deleting from an m-Way Search Tree

Delete the key 84

Search for 84, k2=84 & C1=C2=0, and Simply Delete 84
Data Structures
119
SNU
IDB Lab.
Deleting from an m-Way Search Tree

Delete the key 5 :
(a) Only one key in the node  Need to replace
(b) From C0, move up the element with largest key  move the key 4 to the key 5’s position
Data Structures
120
SNU
IDB Lab.
Deleting from an m-Way Search Tree

Delete the key 10
Replace this element with either the largest element in C0 or smallest element in C1
So, element with key 5 is moved to top & element with key 4 is moved up to the key 5’s position
Data Structures
121
SNU
IDB Lab.
Height of an m-Way Search Tree


h : Height, n : number of elements, m : m-way
The number of elements: h ≤ n ≤ mh – 1
h-1

The number of nodes : ∑
mi = (mh-1)/(m-1) nodes
i=0

The range of height: logm(n+1) ≤ h ≤ n

The number of disk accesses : O(h)

We want to ensure that the height h is close to logm(n+1)  this is
accomplished by B-tree!
Data Structures
122
SNU
IDB Lab.
Table of Contents

B-TREES

Indexed Sequential Access Method(ISAM)
m-WAY Search Trees

B-Trees of Order m





Height of a B-Tree
Searching a B-Tree
Inserting into a B-Tree
Deletion from a B-Tree
Data Structures
123
SNU
IDB Lab.
Definition: B tree of Order m
B-tree is a m-way search tree satisfying the following properties

1.
2.
3.

The root has at least two children
All internal nodes other than the root at least m/2 children (pointers to
the children nodes)
All external nodes are at the same level
Internal node has several pairs of a key and a pointer to a disk block
Data Structures
124
SNU
IDB Lab.
B-Trees of Order m



B-tree of order 2: Fully binary tree
B-tree of order 3 (= 2- 3 tree): 2 or 3 children
B-tree of order 4 (= 2- 3- 4 tree): 2 or 3 or 4 children
Data Structures
125
SNU
IDB Lab.
Table of Contents

B-TREES

Indexed Sequential Access Method(ISAM)
m-WAY Search Trees
B-Trees of Order m

Height of a B-Tree





Searching a B-Tree
Inserting into a B-Tree
Deletion from a B-Tree
Data Structures
126
SNU
IDB Lab.
Height of a B-Tree of Order m

Remember: All internal nodes other than the root at least m/2 children
(pointers to the children nodes)

Lemma 16.3
Let T be a B-tree of order m
Let h be the height of T
Let d= m/2 be the degree of T
Let n be the number of elements in T

(a) 2dh-1 ≤ n ≤ mh – 1

(b) logm(n + 1) ≤ h ≤ logd((n+1)/2) + 1
Data Structures
127
SNU
IDB Lab.
Table of Contents

B-TREES

m-WAY Search Trees
B-Trees of Order m

Height of a B-Tree

Searching a B-Tree



Inserting into a B-Tree
Deletion from a B-Tree
Data Structures
128
SNU
IDB Lab.
Searching a B-Tree

Using the same algorithm as is an m-way search tree





First visit the root with the given key K
Compare K and the keys in the root
Follow the corresponding pointer
Search the child node recursively until the leaf node
If arrived at the leaf node, Search the external node
Data Structures
129
SNU
IDB Lab.
Table of Contents

B-TREES

Indexed Sequential Access Method(ISAM)
m-WAY Search Trees
B-Trees of Order m
Height of a B-Tree
Searching a B-Tree

Inserting into a B-Tree

Deletion from a B-Tree




Data Structures
130
SNU
IDB Lab.
Inserting into a B-Tree



First search with the key of the new element
Found  Insertion fails (if duplicates are not permitted)
Not Found
Insert the new element into the last encountered internal node
If (no overflow) return ok
Else (overflow) { split the last internal node into 2 new nodes;
go to the 1-level up for updating the parent node (recursively)}
Data Structures
131
SNU
IDB Lab.
Notations in B-tree

e : element

Full node has m elements & m+1 children


d : degree of a node  at least m/2

ei : element pointers
d-1, c0, (e1, c1), …, (ed1-1, cd-1)
Q : Right remainder


m, c0, (e1, c1), …, (em, cm)
P : Left remainder


ci : children pointers
Overfull node


c : children p : parent node
m-d, cd, (ed+1, cd+1), …, (em, cm)
Pair(ed, Q) is inserted into the parent of P
Data Structures
132
SNU
IDB Lab.
Insert the key 3 in B-tree
Data Structures
133
SNU
IDB Lab.
Insert the key 25 in B-tree

d = 4 & the target node (“6”, 20,30,40,50,60,70)
 P : 3, 0, (20,0), (25,0), (30,0)
 Q : 3, 0, (50,0), (60,0), (70,0)
 (40, Q) is inserted into parent of P
P
Data Structures
Q
134
SNU
IDB Lab.
Growing B tree by Insertion (1)
30 80
20
10
50 60
25
35 40
55
90
70
82 85
95
Fig 16.25 B-tree of order 3 (at least 2 pointers)
node format: M, C0, (e1, c1), (e2, c2)… (em, cm)
where m= no of elements, ei = elements, ci = children
Data Structures
135
SNU
IDB Lab.
Growing B tree by Insertion (2)
Insert 44
30 80
20
10


50 60
25
35 40 44
55
90
70
82 85
95
d = 2 & the target node was (2, c5, (35,c6),(40, c7))
Overfull node

3, c5, (35,c6), (40,c7), (44,cn)
Data Structures
136
SNU
IDB Lab.
Growing B tree by Insertion (3)
30 80
S
20
10

82 85
95
d= 3/2  2, split the overfull node into P & Q



25
40 50 60
P Q
C D
35 44 55 70
T
90
P : 1, 0, (35,0)
Q : 1, 0, (44,0)
(40,Q) into the parent A of P

Again the parent A is overfull node
Data Structures
137
SNU
IDB Lab.
Growing B tree by Insertion (4)
S
20
10

25
30 80
A
40 50 60
P Q
C D
35 44 55 70
T
90
82 85
95
Node A is again the overfull node

A : 3, P, (40,Q), (50,C), (60,D)
Data Structures
138
SNU
IDB Lab.
Growing B tree by Insertion (5)
S
20
10



T
90
82 85
95
d= 3/2 = 2, split the node A into A & B


25
R
30 50 80
B
A
40
60
P
Q C
D
35 44 55 70
A : 1, P, (40,Q)
B : 1, C, (60,D)
Move (50,B) into the parent of A
Again the parent of A is overfull node
Data Structures
139
SNU
IDB Lab.
Growing B-tree by Insertion (6)
S
20
10

25
R
30 50 80
B
A
40
60
P
Q C
D
35 44 55 70
T
90
82 85
95
The root node R is now the overfull node

R : 3, S, (30,A), (50,B), 80,T)
Data Structures
140
SNU
IDB Lab.
Growing B tree by Insertion (7)
50
R
30
S
20
10



B
60
A
40
P
35
Q C
44 55
D
70
T
90
82 85
95
d= 3/2  2, split the root node R into R & U


25
U
80
R : 1, S, (30,A)
U : 1, B, (80,T)
Move the new index (50, U) into the parent of R
R has no parent, we create a new root for the new index
Data Structures
141
SNU
IDB Lab.
Disk accesses in B tree


Worst case: Insertion may cause s nodes to split upto root
Number of disk accesses in the worst case
h (to read in the nodes on the search path)
+
2s (to write out the two split parts of each node)
+
1 (to write the new root or the node into which an insertion that does not result in a
split is made)

 h + 2s + 1

 at most 3h + 1 because s is at most h

The worst scenario is to have 3h+1 disk IOs by splitting
Data Structures
142
SNU
IDB Lab.
Table of Contents

B-TREES

Indexed Sequential Access Method (ISAM)
m-WAY Search Trees
B-Trees of Order m
Height of a B-Tree
Searching a B-Tree
Inserting into a B-Tree

Deletion from a B-Tree





Data Structures
143
SNU
IDB Lab.
Deletion from a B-Tree
Deletion cases



Case 1: Key k is in the leaf node
Case 2: Key k is in the internal node
Case 2


by replacing the deleted element with



The largest element in its left-neighboring subtree
The smallest element in its right-neighboring subtree
Replacing element is supposed to be in a leaf, so we can apply
case 1
Data Structures
144
SNU
IDB Lab.
Case 1: Leaf Node Deletion

If key k is in leaf node, then remove k from leaf node
X

If underfull node happens, care must be exercised (will address shortly)
Data Structures
145
SNU
IDB Lab.
Case 2: Internal Node Deletion

If the key k is in the internal node x
One of 3 subcases:
a. If the left child y preceding k in x has ≥ t keys
b. If the right child z following k in x has ≥ t keys
c. If both the left and right subchild y and z have t-1 keys

t : m/2 - 1 (half of the keys)
Data Structures
146
SNU
IDB Lab.
Case 2a: Internal Node Deletion (1)
If the left child y preceding k in x has ≥ t keys


Find predecessor k' of k in subtree rooted at y
Replace k by k' in x
x
Data Structures
147
SNU
IDB Lab.
Case 2a: Internal Node Deletion (2)

If underfull node happens, care must be exercised (will address shortly)
Data Structures
148
SNU
IDB Lab.
Case 2b: Internal Node Deletion


If the right child z following k in x has ≥ t keys:
(a) Find successor k' of k in subtree y,
(b) Replace k by k' in x
If underfull node happens, care must be exercised (will address shortly)
Data Structures
149
SNU
IDB Lab.
Case 2c: Internal Node Deletion

If both the left and right subchild y and z have t-1 keys


Select the replacement as shown in case 2a or case 2b
If underfull node happens, care must be exercised as shown in the below
Data Structures
150
SNU
IDB Lab.
Shrinking B-Tree by Deletion (1)
50
R
30
S
20
10
25
U
80
A
40
P
35
Q C
44 55
B
60
* Try to delete “44”
T
90
D
70
82 85
95
50
R
30

S
20
After deleting “44”, “35”
& “40” are merged
10
Data Structures
25
151
U
80
A
40
P
35
Q C
44 55
B
60
D
70
T
90
82 85
SNU
IDB Lab.
95
Shrinking B-Tree by Deletion (2)


“20” & “40” also needs to merged
“50” and “80” also needs to merged
Data Structures
152
SNU
IDB Lab.
Shrinking B-Tree by Deletion (3)


“50” & “80” are merged and now
the old root becomes empty
Free the old root and
make the new root
Data Structures
153
SNU
IDB Lab.
Technique for Reducing Node Merging
: B tree Deletion with Redistribution (1)
Try to delete “25”
Underflow happens & Redistribute some neighbor nodes


Move down 10 & move up 6
Data Structures
Save node merging
154
SNU
IDB Lab.
Technique for Reducing Node Merging
: B tree Deletion with Redistribution (2)
Try to delete “10”
Merging is unavoidable
Data Structures
155
SNU
IDB Lab.
Technique for Reducing Node Merging
: B tree Deletion with Redistribution (3)

Consider redistributing some nodes: move down “30” & move up “50”
Data Structures
Save propagation of
156 node merging
SNU
IDB Lab.
Summary (0)

Chapter 15: Binary Search Tree


Chapter 16: Balanced Search Tree



BST and Indexed BST
AVL tree: BST + Balance
B-tree: generalized AVL tree
Chapter 17: Graph
Data Structures
157
SNU
IDB Lab.
Summary (1)

Balanced tree structures
- Height is O(log n)

AVL and Red-black trees


Splay trees



Suitable for internal memory applications
Individual dictionary operation  0(n)
Take less time to perform a sequence of u operations  0(u log u)
B-trees

Suitable for external memory
Data Structures
158
SNU
IDB Lab.