Lecture3

Advance Data Structures

Lecture 3 AVL Trees

2

Binary Search Tree - Best Time

• All BST operations are O(h), where h is tree height

• minimum h is for a binary tree with N nodes› What is the best case tree? › What is the worst case tree?

• So, best case running time of BST operations is O(log N)

Nlogh 2

3

Binary Search Tree - Worst Time

• Worst case running time is O(N) › What happens when you Insert elements in

ascending order?• Insert: 2, 4, 6, 8, 10, 12 into an empty BST

› Problem: Lack of “balance”: • compare height of left and right subtree

› Unbalanced degenerate tree

4

Balanced and unbalanced BST

4

2 5

1 3

1

5

2

4

3

7

6

4

2 6

5 71 3

Is this “balanced”?

5

Approaches to balancing trees• Don't balance

› May end up with some nodes very deep

• Strict balance› The tree must always be balanced perfectly

• Pretty good balance› Only allow a little out of balance

• Adjust on access› Self-adjusting

6

Balancing Binary Search Trees

• Many algorithms exist for keeping binary search trees balanced› Adelson-Velskii and Landis (AVL) trees

(height-balanced trees) › Splay trees and other self-adjusting trees› B-trees and other multiway search trees

7

Perfect Balance• Want a complete tree after every operation

› tree is full except possibly in the lower right• This is expensive

› For example, insert 2 in the tree on the left and then rebuild as a complete tree

Insert 2 &complete tree

6

4 9

81 5

5

2 8

6 91 4

8

AVL - Good but not Perfect Balance

• AVL trees are height-balanced binary search trees

• Balance factor of a node› height(left subtree) - height(right subtree)

• An AVL tree has balance factor calculated at every node› For every node, heights of left and right

subtree can differ by no more than 1› Store current heights in each node

9

Height of an AVL Tree

• N(h) = minimum number of nodes in an AVL tree of height h.

• Basis› N(0) = 1, N(1) = 2

• Induction› N(h) = N(h-1) + N(h-2) + 1

• Solution (recall Fibonacci analysis)

› N(h) > h ( 1.62) h-1h-2

h

10

Height of an AVL Tree

• N(h) > h ( 1.62)• Suppose we have n nodes in an AVL

tree of height h.› n > N(h) (because N(h) was the minimum)

› n > h hence log n > h (relatively well balanced tree!!)

› h < 1.44 log2n (i.e., Find takes O(logn))

11

Node Heights

1

00

2

0

6

4 9

81 5

1

height of node = hbalance factor = hleft-hright

empty height = -1

0

0

height=2 BF=1-0=1

0

6

4 9

1 5

1

Tree A (AVL) Tree B (AVL)

12

Node Heights after Insert 7

2

10

3

0

6

4 9

81 5

1

height of node = hbalance factor = hleft-hright

empty height = -1

1

0

2

0

6

4 9

1 5

1

0

7

0

7

balance factor 1-(-1) = 2

-1

Tree A (AVL) Tree B (not AVL)

13

Insert and Rotation in AVL Trees

• Insert operation may cause balance factor to become 2 or –2 for some node › only nodes on the path from insertion point to

root node have possibly changed in height› So after the Insert, go back up to the root

node by node, updating heights› If a new balance factor (the difference hleft-

hright) is 2 or –2, adjust tree by rotation around the node

14

Single Rotation in an AVL Tree

2

10

2

0

6

4 9

81 5

1

0

7

0

1

0

2

0

6

4

9

8

1 5

1

0

7

15

Let the node that needs rebalancing be .

There are 4 cases: Outside Cases (require single rotation) : 1. Insertion into left subtree of left child of . 2. Insertion into right subtree of right child of . Inside Cases (require double rotation) : 3. Insertion into right subtree of left child of . 4. Insertion into left subtree of right child of .

The rebalancing is performed through four separate rotation algorithms.

Insertions in AVL Trees

16

j

k

X Y

Z

Consider a validAVL subtree

AVL Insertion: Outside Case

h

hh

17

j

k

XY

Z

Inserting into Xdestroys the AVL property at node j

AVL Insertion: Outside Case

h

h+1 h

18

j

k

XY

Z

Do a “right rotation”

AVL Insertion: Outside Case

h

h+1 h

19

j

k

XY

Z

Do a “right rotation”

Single right rotation

h

h+1 h

20

j

k

X Y Z

“Right rotation” done!(“Left rotation” is mirror symmetric)

Outside Case Completed

AVL property has been restored!

h

h+1

h

21

j

k

X Y

Z

AVL Insertion: Inside Case

Consider a validAVL subtree

h

hh

22

Inserting into Y destroys theAVL propertyat node j

j

k

XY

Z

AVL Insertion: Inside Case

Does “right rotation”restore balance?

h

h+1h

23

jk

X

YZ

“Right rotation”does not restorebalance… now k isout of balance

AVL Insertion: Inside Case

hh+1

h

24

Consider the structureof subtree Y… j

k

XY

Z

AVL Insertion: Inside Case

h

h+1h

25

j

k

X

V

Z

W

i

Y = node i andsubtrees V and W

AVL Insertion: Inside Case

h

h+1h

h or h-1

26

j

k

X

V

Z

W

i

AVL Insertion: Inside Case

We will do a left-right “double rotation” . . .

27

j

k

X V

ZW

i

Double rotation : first rotation

left rotation complete

28

j

k

X V

ZW

i

Double rotation : second rotation

Now do a right rotation

29

jk

X V ZW

i

Double rotation : second rotation

right rotation complete

Balance has been restored

hh h or h-1

30

Implementation

balance (1,0,-1)

key

rightleft

No need to keep the height; just the difference in height, i.e. the balance factor; this has to be modified on the path of insertion even if you don’t perform rotations

Once you have performed a rotation (single or double) you won’t need to go back up the tree

31

Single Rotation

RotateFromRight(n : reference node pointer) {p : node pointer;p := n.right;n.right := p.left;p.left := n;n := p}

X

Y Z

n

You also need to modify the heights or balance factors of n and p

Insert

32

Double Rotation

• Implement Double Rotation in two lines.

DoubleRotateFromRight(n : reference node pointer) {????}

X

n

V W

Z

33

Insertion in AVL Trees

• Insert at the leaf (as for all BST)› only nodes on the path from insertion point to

root node have possibly changed in height› So after the Insert, go back up to the root

node by node, updating heights› If a new balance factor (the difference hleft-

hright) is 2 or –2, adjust tree by rotation around the node

34

Insert in BST

Insert(T : reference tree pointer, x : element) : integer {if T = null then T := new tree; T.data := x; return 1;//the links to //children are nullcase T.data = x : return 0; //Duplicate do nothing T.data > x : return Insert(T.left, x); T.data < x : return Insert(T.right, x);endcase}

35

Insert in AVL trees

Insert(T : reference tree pointer, x : element) : {if T = null then {T := new tree; T.data := x; height := 0; return;}case T.data = x : return ; //Duplicate do nothing T.data > x : Insert(T.left, x); if ((height(T.left)- height(T.right)) = 2){ if (T.left.data > x ) then //outside case T = RotatefromLeft (T); else //inside case T = DoubleRotatefromLeft (T);} T.data < x : Insert(T.right, x); code similar to the left caseEndcase T.height := max(height(T.left),height(T.right)) +1; return;}

36

Example of Insertions in an AVL Tree

1

0

2

20

10 30

25

0

35

0

Insert 5, 40

37

Example of Insertions in an AVL Tree

1

0

2

20

10 30

25

1

35

0

50

20

10 30

25

1

355

40

0

0

01

2

3

Now Insert 45

38

Single rotation (outside case)

2

0

3

20

10 30

25

1

35

2

50

20

10 30

25

1

405

40

0

0

0

1

2

3

45

Imbalance35 45

0 0

1

Now Insert 34

39

Double rotation (inside case)

3

0

3

20

10 30

25

1

40

2

50

20

10 35

30

1

405

45

0 1

2

3

Imbalance

45

0

1

Insertion of 34

35

34

0

0

1 25 340

Extended Example

Insert 3,2,1,4,5,6,7, 16,15,14

41

AVL Tree Deletion

• Similar but more complex than insertion› Rotations and double rotations needed to

rebalance› Imbalance may propagate upward so that

many rotations may be needed.

42

Deletion of a Node

• Deletion of a node x from an AVL tree requires the same basic ideas, including single and double rotations, that are used for insertion.

• With each node of the AVL tree is associated a balance factor that is left high, equal or right high according, respectively, as the left subtree has height greater than, equal to, or less than that of the right subtree.

43

Method

1. Reduce the problem to the case when the node x to be deleted has at most one child.

› If x has two children replace it with its immediate predecessor y under inorder traversal (the immediate successor would be just as good)

› Delete y from its original position, by proceeding as follows, using y in place of x in each of the following steps.

44

Method (cont.)

2. Delete the node x from the tree.› We’ll trace the effects of this change on height through all the

nodes on the path from x back to the root.› The action to be taken at each node depends on

• balance factor of the node• sometimes the balance factor of a child of the node.

45

Rebalancing the tree

Let p be the node whose balance factor becomes 2 or -2.› Rotation is needed.› Let q be the root of the taller subtree of p. We

have three cases according to the balance factor of q.

46

Case a

Case a: The balance factor of q is 0.› Apply a single rotation› Height of p is unchanged.

0 q

T3T2deleted

T1

-1 p

h-1

h h

1 q

T3

T2T1

-1 p

h-1h

h

height unchanged

47

Case bCase b: The balance factor of q is the same as that of

the previous balance factor of p.› Apply a single rotation› Set the balance factors of p and q to 0› Height of the subtree reduced .

-1q

T3T2deleted

T1

-1 p

h-1

h-1h

0q

T3

T1

0p

h-1h

height reduced

T2

h-1

48

Case cCase c: The previous balance factors of p and balance

factor of q are opposite.› Apply a double rotation› set the balance factors of the new root to 0› Height of subtree is reduced. height reduced

h-1T4

h-1q

T3T1

r

h-1T2

h-1 or h-2

p

0

1 q

T3

T1

r

h-1

T2

h-1 or h-2

p-1

T4

Delete 55

60

20 70

10 40 65 85

5 15 30 50 80 90

55

Delete 55

60

20 70

10 40 65 85

5 15 30 50 80 90

55

Delete 50

60

20 70

10 40 65 85

5 15 30 50 80 90

55

Delete 50

60

20 70

10 40 65 85

5 15 3050 80 90

55

Delete 60

60

20 70

10 40 65 85

5 15 30 50 80 90

55

prev

Delete 60

55

20 70

10 40 65 85

5 15 30 50 80 90

Delete 40

40

20 70

10 30 65 85

5 15 80 90

Delete 40

40

20 70

10 30 65 85

5 15 80 90

prev

Delete 40 : Rebalancing

30

20 70

10 65 85

5 15 80 90

Case ?

Delete 40: after rebalancing

30

7010

20 65 855

15 80 90

Single rotation is preferred!

59

Delete 20

30

20 55

10 25 45 60

15 5950

57

65

60

Delete 20

30

25 55

10 45 60

15 5950

57

65

61

Delete 20

30

15 55

10 45 6025

5950

57

65

62

Delete 20

55

30 60

15 59 6545

10 25 5750

63

Arguments for AVL trees:

1. Search is O(log N) since AVL trees are always balanced.2. Insertion and deletions are also O(logn)3. The height balancing adds no more than a constant factor to the

speed of insertion.

Arguments against using AVL trees:4. Difficult to program & debug; more space for balance factor.5. Asymptotically faster but rebalancing costs time.6. Most large searches are done in database systems on disk and use

other structures (e.g. B-trees).7. May be OK to have O(N) for a single operation if total run time for

many consecutive operations is fast (e.g. Splay trees).

Pros and Cons of AVL Trees

64

Double Rotation Solution

DoubleRotateFromRight(n : reference node pointer) {RotateFromLeft(n.right);RotateFromRight(n);}

X

n

V W

Z