CS3424
AVL Trees
Red-Black Trees

Trees
• As stated before, trees are great ways of holding hierarchical data
• Insert, Search, Delete ~ O(lgN)
• But only if they’re balanced!
• So let’s discuss how to assure balance

Rotations
 x
 y

Left-Rotate(x)
Right-Rotate(y)
 x
 y


AVL Trees (1962)
• Uses height property to maintain balance
• Height of left & right differ by at most 1
• Uses rotations to restore balance after insertions and deletions

Balancing in AVL Trees
• Balance is the difference between the heights of the left and right subtrees
– This should range between -1 and 1
• Maintain this balance upon insert and delete
• Balance = H
R
- H
L


Left-Left Imbalance
Right-Rotate(X)
-1
Y
-2
X



Y
0
X
0




Left-Right Imbalance
+1
Y
-2
X

=

+1
Y
-1

-2
X




Left-Right Imbalance (cont)
+1
Y
-2
X

=

+1
Y
+1

-2
X



Left-Right Imbalance (
 left)

+1
Y
-1

-2
X


Left-Right Imbalance (
 left)

+1
Y
-1

-2
X

Left-Rotate(Y)
Y
-2

0
-2
X




Left-Right Imbalance (
 left)
-2

Y
0
-2
X

Right-Rotate(X)
0
Y


0
X
+1


Left-Right Imbalance (
 right)

+1
Y
+1

-2
X


Left-Right Imbalance (
 right)

+1
Y
+1

-2
X

Left-Rotate(Y)
-1
Y
-1

-2
X



Left-Right Imbalance (
 right)

-1
Y
-1

-2
X

Right-Rotate(X)
-1
Y

0 
X
0


Red-Black Trees (1972/1978)
• BST + notion of color (RED / BLACK)
• Every node is either red or black
• Root is black
• Every leaf (NIL) is black
• If node red, both children are black
• For each node, path to NIL children contains same number of black nodes ( black-height )

Possibilities in Growing a Red-
Black Tree

Painting a Tree to be
Red-Black

Painting a Tree to be
Red-Black
Root is black

Painting a Tree to be
Red-Black
Must be black bh=1 bh=1

Painting a Tree to be
Red-Black

Painting a Tree to be
Red-Black

Painting a Tree to be
Red-Black
Must be black

Painting a Tree to be
Red-Black
Must be red

Painting a Tree to be
Red-Black bh = 2 or 3

Inserting & Deleting in Red-
Black Trees
• Insert similar to binary search trees
– “Search” left & right until finding NIL
– But now, we need to worry about coloring
– Requires possible rotations & “cleanup”
• Deleting similar to BSTs as well
– Search left & right until finding item
– If we removed a black node, “cleanup”

Red-Black Insert
• Insert node (z)
• z.Color = red
– This can cause two possible problems:
• Root must be black (initial insert violates this)
• Two red nodes cannot be adjacent (this is violated if parent of z is red)
– Cleanup

Red-Black Cleanup
• y = z’s “uncle”
• Three cases:
– y is red
– y is black and z is a right child
– y is black and z is a left child
Not mutually exclusive

Case 1 – z’s Uncle is red
• y.Color = black
• z.Parent.Color = black
• z.Parent.Parent.Color = red
• z = z.Parent.Parent
• Repeat cleanup

Case 1 Visualized
11
2
1 7
5 8 y z
4
New
Node
14
15 y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent
repeat cleanup

Case 1 Visualized
11
2
1 7
5 8 y z
4
New
Node
14
15 y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent
repeat cleanup

Case 1 Visualized
11
2
1 7
5 8 y z
4
New
Node
14
15 y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent
repeat cleanup

Case 1 Visualized
11
2
1 7
5 8 y z
4
New
Node
14
15 y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent
repeat cleanup

Case 1 Visualized
11
2
1 z
7
5 8
4
New
Node
14 y
15 y.Color = black z.Parent.Color = black z.Parent.Parent.Color = red z = z.Parent.Parent
repeat cleanup

Case 2 – z’s Uncle is black & z is a right child
• z = z.Parent
• Left-Rotate(T, z)
• Do Case 3
• Note that Case 2 is a subset of Case 3

Case 2 Visualized
11
2
1 z
7
5 8
4
14 y
15 z = z.Parent
Left-Rotate(T,z)
Do Case 3

Case 2 Visualized
11 z 2
1 7
5 8
4
14 y
15 z = z.Parent
Left-Rotate(T,z)
Do Case 3

Case 2 Visualized
11
7
1 z 2
5
8
4
14 y
15 z = z.Parent
Left-Rotate(T,z)
Do Case 3

Case 3 – z’s Uncle is black & z is a left child
• z.Parent.Color = black
• z.Parent.Parent.Color = red
• Right-Rotate(T, z.Parent.Parent)

Case 3 Visualized
11
7
1 z 2
5
8
4
14 y
15 z.Parent.Color = black z.Parent.Parent.Color = red
Right-Rotate(T, z.Parent.Parent)

Case 3 Visualized
11
7
1 z 2
5
8
4
14 y
15 z.Parent.Color = black z.Parent.Parent.Color = red
Right-Rotate(T, z.Parent.Parent)

Case 3 Visualized
11
7
1 z 2
5
8
4
14 y
15 z.Parent.Color = black z.Parent.Parent.Color = red
Right-Rotate(T, z.Parent.Parent)

Case 3 Visualized
7
1 z 2
5
4
11
8 14 y
15 z.Parent.Color = black z.Parent.Parent.Color = red
Right-Rotate(T, z.Parent.Parent)

Analysis of Red-Black Cleanup

C New z C
A
 z B


D y

A D

B
 
 
Case 1

,

,

,

, and
 are all black-rooted and all have same black-height
(symmetric L/R on A & B)

Analysis of Red-Black Cleanup

C
A
 z B

 y
Case 2
C
B
 y

A z

 Case 3
Case 2

,

,

, and
 are all black-rooted and all have same black-height
(
 is black or this would be case 1)

Analysis of Red-Black Cleanup
C B
 y

A z

B


A z
 
C
 y
Case 3

,

,

, and
 are all black-rooted and all have same black-height

Questions about
Red-Black Insert
• What is the running time?
• Why are there only at most 2 rotations in the cleanup?
• How many times can the cleanup repeat (i.e. how many times can we have case 1 in a row)?

Red-Black Delete
• Delete as before
• But if you take away a black node:
– Black height is not “balanced”
– Can generate adjacent red nodes
– Must perform cleanup
• 4 cases
• “Beyond the scope of this course…”

FIN