Data Structure: Chapter 8
Min Chen
School of Computer Science and Engineering
Seoul National University

Motivation of Red-Black Trees
 Balanced and Unbalanced Trees
 AVL Trees


Definition of Red-Black Trees
Operations for Red-Black Trees
 Search
 Insertion

Balanced and Unbalanced Trees
20
10
5
30
15
25
35

AVL-Tree
Balance Factor:
0-1=-1
Unbalanced
 The balance factor of a
node is the height of its
left subtree minus the
height of its right
subtree
 A node with balance
factor 1, 0, or -1 is
considered balanced
60
Balance Factor:
2-0=2
40
50
30
45

AVL-Tree: Rotation Operation

AVL-Tree: Rotation Operation
60
Left Right Case
50
40
40
60
60
50
45
45
30
30
30
45

Definition: a binary tree, satisfying:
1. Every node is colored either red or black
2. The root is black
3. If a node is red, its children must be black
▪ consecutive red nodes are disallowed
4. Every path from a node to a null reference must
contain the same number of black nodes

The insertion sequence is
 10, 85, 15, 70, 20, 60, 30, 50, 65, 80, 90, 40, 5, 55
30
15
10
70
20
60
5
50
40
85
65
55
80
90
• Each path must contain the same number of black nodes. (Rule
#4)
• Consecutive red nodes are not allowed. (Rule #3)
• The longest path is at most twice the length of the shortest
path
2 B  1  N  22 B  1
2 B  N  1  22 B
log 2 B  B
log 2 2 B  2 B
B  log( N  1) log( N  1)  2 B
1
log( N  1)  B  log( N  1)
2
log( N  1)  H  2 log( N  1)



B = total black
nodes from root
to leaf
N = total all nodes
H = height
All operations guaranteed
logarithmic!

A new node must be colored red
 Why?
▪ A new item is always inserted as a leaf in the tree
▪ If we color a new item black, then the number of black
nodes from root would be different (violate property #4)
 If the parent is black, no problem.
 If the parent is red, we create two consecutive red
nodes (violate property #3)
▪ Thus, we have to do some rotating/recolouring…

X: new node
P: parent
S: sibling
G: Grandparent
Case after insertion:
 Consecutive red (P & X)
 Sibling of parent (S) is black
 X is “outer node” (left-left or right-right)
P
G
P
X
A
C
B
X
S
D
E
A
G
B
C
S
D
E

X: new node
P: parent
S: sibling
G: Grandparent
Case after insertion:
 Consecutive red (P & X)
 Sibling of parent (S) is black
 X is “inner node” (left-right or left)
X
G
P
X
A
B
P
S
D
C
E
A
G
B
C
S
D
E
Case after insertion:

 Consecutive red
 Sibling of parent is red
 Outer node (left-left or right-right)
P
G
P
X
A
C
B
X
S
D
E
A
G
B
C
S
D
But what if P’s parent is red?  We have to keep going up
the tree all the way to the root 
E


The solution: prevent S from ever being red!
Starting from the root (searching for
insertion point)
 Never allow 2 red siblings
 If we see a node X with 2 red children, do a colour
flip.
X
C1
X
C2
C1
C2
X
C1
X
C2
C1
C2
 Maintains property #4
 Possible violation of #3: if X’s parent is red!
▪ Do single or double rotation
▪ X’s parent’s sibling can never be red!
 Set the root to black (to maintain property #2)
If we do the colour flipping on the way
down to the insertion point, we will never
reach a condition where P & S are red!
G
G
P
X
A
S
C
B
D
P
E
X
A
S
C
B
D
E
30
15
10
5
70
20
60
18
50
40
85
65
55
80
90
30
15
10
5
2
70
20
60
18
50
40
85
65
55
80
90
30
15
5
2
70
20
10
60
18
50
40
85
65
55
80
90
30
15
5
2
70
20
10
60
18
50
40
65
55
45
85
80
90
30
15
5
2
70
20
10
60
18
Color-flip!
50
40
85
65
80
55
45
90
30
15
5
2
70
20
10
60
18
Color-flip!
50
40
85
65
80
55
45
90
30
15
5
2
70
20
10
60
18
50
40
85
65
80
55
45
90
30
15
5
2
60
50
20
10
18
40
70
55
85
65
80
45
90

The insertion sequence is 10, 85, 15, 70, 20,
60, 30, 50, 65, 80, 90, 40, 5, 55
Deletion in BST: only leaf nodes or nodes with one
child are really deleted (Why?)
 If the deleted node is red: no problem (all properties
maintained).

Leaf nodes:
Single child nodes:



If node to be deleted is black  violate property #4
Always ensure that the node to be deleted is red.
Top-down traversal from root (looking for node to be
deleted):
P
X
A
X: visited node
P: parent
S: sibling
S
B
C
D
Idea: make sure that X is red!


P is red (inductive invariant)
X and S are black (result of property #3)
P
X
A

S
B
C
D
2 cases:
 1. Both X’s children (A & B) are black
 2. X has at least one red child (A, B, or both)

Depends on children of S (C & D):
 Both C & D are black: simply colour-flip:
P
P
X
A
S
B
C
X
D
A
S
B
C
D
Case 1: Both X’s children are black

Outer child of S (C) is Red: do a single rotation and
recolour
S
P
P
X
A
S
B
C
C
X
D
A
D
B

Inner child of S (C) is Red: do a double rotation
and recolour
C
P
X
A
P
S
B
C
X
D
A
S
D
B



Recurse down to X’s child
If we land on a red node, fine.
If we land on a black node, rotate sibling and
parent:
S
P
P
X
A
S
B
C
C
X
D
A
D
B
Red-Black trees use color as balancing
information instead of height in AVL trees.
 An insertion may cause a local perturbation (two
consecutive red nodes)
 The pertubation is either

 resolved locally (rotations), or
 propagated to a higher level in the tree by recoloring
(color flip)




O(1) for a rotation or color flip
At most one restructuring per insertion.
O(log n) color flips
Total time: O(log n)