CHAPTER 4
Trees
§4 AVL Trees
Target : Speed up searching (with insertion and
deletion)
root
Tool : Binary search trees
small
large
Problem : Although Tp = O( height ), but the
height can be as bad as O( N ).
1/19
§4 AVL Trees
〖Example〗 2 binary search trees obtained for the months
of the year
Jan
Feb
Mar
Apr
June
Aug
May
July
Sept
Dec
Oct
Entered from Jan to Dec
Nov
July
Feb
Aug
Apr
May
Jan
Dec
Mar
June
A balanced tree
2/19
Oct
Nov
Sept
Discussion 1:
What are the
average search
times of these
two trees?
What if the
months are
entered in
alphabetical
order?
§4 AVL Trees
Adelson-Velskii-Landis (AVL) Trees (1962)
【Definition】An empty binary tree is height balanced. If T is a
nonempty binary tree with TL and TR as its left and right
subtrees, then T is height balanced iff
(1) TL and TR are height balanced,
and of an empty tree
The height
(2) | hL  hR |  1 where hL and hisR defined
are thetoheights
be –1. of TL and TR ,
respectively.
【Definition】The balance factor BF( node ) = hL  hR . In an
AVL tree, BF( node ) = 1, 0, or 1.
4
3
2
1
3/19
5
5
2
6
1
7
8
4 7
3
7
2
1
8
4
3
5
§4 AVL Trees
〖Example〗 Input the months
Mar
May
1
2
0
Mar
Nov
1
2
01
0
1
0
May
Single rotation
May
0
00
00
Mar
Nov
Nov
 The trouble maker Nov is in the right subtree’s right
subtree of the trouble finder Mar. Hence it is called an
RR rotation.
A is not necessarily
In general:
the root of the tree
1
A
RR
0
4/19
1
BR
B
BL
0
RR
BR
B
A
Rotation
AL
AL
BL
A
Insertion
B
0
2
BR
AL
BL
§4 AVL Trees
Aug
Apr
2
1
01
12
May
210
00
Mar
Nov
10
Aug
0
Apr
Discussion 2: What can we do now?
5/19
§4 AVL Trees
Jan
21
2
1
01
Double Rotation
1
2
01
0
May
1
010
00
Aug
1
010
Nov
0
10
Apr
Mar
Aug
LR
0
Jan
Mar
1
0
May
0
0
Apr
Jan
0
Nov
Rotation
In general:
1
0
B
A
CL
Insertion
1
B
0
C
BL
2
LR
A
BL
CL
6/19
0 or 1
C
1 or 0
B
AR
CR
OR
0
Rotation
1
C
AR
CR
LR
BL
A
CL
CR
OR
AR
§4 AVL Trees
Dec
July
Feb
1
2
01
21
1
2
10
Mar
Aug
May
0
01
Apr
Jan
1
0
1
0
0
0
Dec
Nov
July
0
Feb
Discussion 3: What can we do now?
7/19
June
Oct
§4 AVL Trees
Sept
1
0
1
0 2
1
01
21
Jan
Mar
Jan
1
1
12
1
0 1
00
02
1
0
1
1
12
1
0
0 1
Dec
Mar
May
Dec Dec
Mar
1 10
1 Aug
Aug
Aug
0 0
0 Apr
Apr
Apr
0
01
10 1
1
Jan
2
10
1
0
Feb
July
May
Nov
Feb0Home
July work:
Nov
1
00
1
0 1
Feb
0 July
June 0 Nov 0
Oct
0
p.136June
4.16 May
0
0
June
Oct
Create an AVL Tree
Sept
0
Note: Several bf’s
might be changed
if
p.136even
4.22
we don’t need to reconstruct
Double Rotation
the tree.
Another option is to keep a height field for each node.
Read the declaration and functions in Figures 4.42 – 4.48
8/19
§4 AVL Trees
Let nh be the minimum number of nodes in a height balanced tree of
height h. Then the tree must look like
A
h2
A
h1
last question:
OROne
h1
h2  nh = nh1 + nh2 + 1
Obviously we have Tp = O( h )
where h is the height of the tree.
But h = ?
Fibonacci numbers:
F0 = 0, F1 = 1, Fi = Fi1 + Fi2 for i > 1
 nh = Fh+2  1, for h  0
i


1  1 5 
Fibonacci number theory gives that Fi 


1  1  5 
 nh 
5  2 
9/19
5
h 2
1
 h  O(ln n)
2

§5 Splay Trees
Target : Any M consecutive tree operations
starting from an empty tree take at most
O(M log N) time.
ButDoes
node
takes
O(N)
time
Soif aone
single
operation
might
it mean
that every
Sure
we
can
–
that
only
means
The bound is weaker.
to access,
wetake
can
keepN)
accessing
still
O(N)
time?
operation
takes
O(log
time? it
No.
It
means
that
the
that
whenever
a
node
is
accessed,
But the effect is the same:
for
M times,
we?
Then
what’scan’t
the point?
amortized
time
is
O(log
N).
it
must
be
moved.
There are no bad input sequences.
Discussion 4: How shall we move the node?
10/19
§5 Splay Trees
k1
k2
A
A k2 A
A
A
k4
k4
k3 k1
k3 k4
B k1
k2 k2
D
k1
k3
k2
k3D
k1
k3
B
C
C
D
B
C
B
C C
B
k5
k4 k5
E
k4
E
k5
F
E F
E
E
D
Does NOT work!
11/19
k5
k5
D
F
F
F
§5 Splay Trees
An even worse case:
Discussion 5:
Try to Insert 1, 2, ..., N in increasing order, and
then Find them in the same order. What will
happen? What is the total access time T(N)?
12/19
§5 Splay Trees
Try again -- For any nonroot node X , denote its parent
by P and grandparent by G :
Case 1: P is the root
Rotate X and P
Case 2: P is not the root
Zig-zag
G
X
Double rotation
P
D
A
B
B
G
P
X
D
X
D
C
B
C
C
Zig-zig
13/19
G
X
A
A
P
P
A
Single rotation
G
B
C
D
§5 Splay Trees
Splaying not only moves the accessed
node to the root, but also roughly halves
k5
the depth of most nodes on the path.
k4
F
k13
E
k2
AA
B
14/19
D
k1
B
Ck3
C
C
k3
k4
D
D
E
k5
F
§5 Splay Trees
Insert: 1, 2, 3, 4, 5, 6, 7
Find: 1
7
7
6
6
5
5
4
3
2
1
7
6
6
1
4
4
4
1
2
2
1
2
5
7
5
3
3
3
Read the 32-node example
given in Figures 4.52 – 4.60
15/19
§5 Splay Trees
Deletions:
 Step 1: Find X ;
Home work:
p.136 4.23
Discussion 6: Then what must we do? Please
Accessdescription
a splay tree
complete the algorithm
for Deletion.
Are splay trees really better
than AVL trees?
16/19
 Other Operations on Binary Search Trees
 Sort: List the elements in increasing order
Solution: inorder traversal.
 Get Height: Compute the heights of the nodes
Solution: postorder traversal.
 Get Depth: Compute the depths of the nodes
Solution: preorder traversal.
17/19
Research Project 1
Binary Search Trees (25)
This project requires you to implement operations on
unbalanced binary search trees, AVL trees, and splay trees.
You are to analyze and compare the performances of a
sequence of insertions and deletions on these search tree
structures.
Detailed requirements can be downloaded from
http://acm.zju.edu.cn/dsaa/
18/19
Research Project 2
Population (25)
It is always exciting to see people settling in a new continent. As
the head of the population management office, you are supposed to
know, at any time, how people are distributed in this continent.
The continent is divided into square regions, each has a center
with integer coordinates (x, y). Hence all the people coming into that
region are considered to be settled at the center position. Given the
positions of the corners of a rectangle region, you are supposed to
count the number of people living in that region.
Note: there are up to 32768 different regions and possibly even
more queries.
Detailed requirements can be downloaded from
http://acm.zju.edu.cn/dsaa/
19/19