The document discusses AVL trees, which are self-balancing binary search trees. It provides information on AVL tree operations like insertion and deletion of nodes. Insertion may cause imbalance, requiring rotation operations like single, double, left, or right rotations to rebalance the tree. Deletion is similar but can require propagating rotations upward to restore balance. AVL trees provide O(log n) time for operations by staying balanced through rebalancing rotations after inserts and deletes.

1. AVL Trees

2. Binary Search Tree - Best Time
• All BST operations are O(h), where h is tree
height
• minimum h is O(log n) for a binary tree with
N nodes
– What is the best case tree? O(log n)
– What is the worst case tree? O(n)
• So, best case running time of BST operations
is O(log N)
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 depths of left and right subtree
– Unbalanced degenerate tree
3

4. Balanced and unbalanced BST
4
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
5

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
6

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
7
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 balance in each node
8

9. Implementation
9
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

10. 5
3 8
1 4 10
5
3
1 4
AVL Tree
Not AVL Tree

11. Node Heights
11
1
00
2
0
6
4 9
81 5
1
height of node = h
balance 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
12
2
10
3
0
6
4 9
81 5
1
height of node = h
balance 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
13

14. Inserting a Node in an AVL Tree
• During insertion, the new node is inserted as the leaf node, so it will
always have balance factor equal to zero.
• The nodes whose balance factors will change are those which lie on
the path between the root of the tree and the newly inserted node.
• The possible changes which may take place in any node on the path
are as follows:
 Initially the node was either left or right heavy and after insertion has
become balanced.
 Initially the node was balanced and after insertion has become either
left or right heavy.
 Initially the node was heavy (either left or right) and the new node has
been inserted in the heavy sub-tree thereby creating an unbalanced
sub-tree. Such a node is said to be a critical node.

15. Single Rotation in an AVL Tree
15
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

16. 5
3
1 4
Insert 0.8
AVL Tree
8
0.8
5
3
1 4
8 Z
3
5
1
0.8
4 8
After Rotation

17. 17
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

18. Rotations to Balance AVL Trees
• To perform rotation, our first work is to find the critical node. Critical node is
the nearest ancestor node on the path from the root to the inserted node
whose balance factor is neither -1, 0 nor 1.
• The second task is to determine which type of rotation has to be done.
• There are four types of rebalancing rotations and their application depends on
the position of the inserted node with reference to the critical node.
 LL rotation: the new node is inserted in the left sub-tree of the left sub-tree of
the critical node
 RR rotation: the new node is inserted in the right sub-tree of the right sub-tree
of the critical node
 LR rotation: the new node is inserted in the right sub-tree of the left sub-tree
of the critical node
 RL rotation: the new node is inserted in the left sub-tree of the right sub-tree
of the critical node

19. 19
j
k
X Y
Z
Consider a valid
AVL subtree
AVL Insertion: Outside Case
h
h
h

20. 20
j
k
X
Y
Z
Inserting into X
destroys the AVL
property at node j
AVL Insertion: Outside Case
h
h+1 h

21. 21
j
k
X
Y
Z
Do a “right rotation”
AVL Insertion: Outside Case
h
h+1 h

22. 22
j
k
X
Y
Z
Do a “right rotation”
Single right rotation
h
h+1 h

23. 23
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

24. 24
j
k
X Y
Z
AVL Insertion: Inside Case
Consider a valid
AVL subtree
h
hh

25. 25
Inserting into Y
destroys the
AVL property
at node j
j
k
X
Y
Z
AVL Insertion: Inside Case
Does “right rotation”
restore balance?
h
h+1h

26. 26
j
k
X
Y
Z
“Right rotation”
does not restore
balance… now k is
out of balance
AVL Insertion: Inside Case
h
h+1
h

27. 27
Consider the structure
of subtree Y… j
k
X
Y
Z
AVL Insertion: Inside Case
h
h+1h

28. 28
j
k
X
V
Z
W
i
Y = node i and
subtrees V and W
AVL Insertion: Inside Case
h
h+1h
h or h-1

29. 29
j
k
X
V
Z
W
i
AVL Insertion: Inside Case
We will do a left-right
“double rotation” . . .

30. 30
j
k
X V
Z
W
i
Double rotation : first rotation
left rotation complete

31. 31
j
k
X V
Z
W
i
Double rotation : second
rotation
Now do a right rotation

32. 32
jk
X V ZW
i
Double rotation : second
rotation
right rotation complete
Balance has been
restored
hh h or h-1

33. 5
3
1 4
Insert 3.5
AVL Tree
8
3.5
5
3
1 4
8
4
5
0.8
3
3.5
After Rotation
8

34. 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

35. Example of Insertions in an AVL Tree
35
1
0
2
20
10 30
25
0
35
0
Insert 5, 40

36. Example of Insertions in an AVL Tree
36
1
0
2
20
10 30
25
1
35
0
5
0
20
10 30
25
1
355
40
0
0
0
1
2
3
Now Insert 45

37. Single rotation (outside case)
37
2
0
3
20
10 30
25
1
35
2
5
0
20
10 30
25
1
405
40
0
0
0
1
2
3
45
Imbalance
35 45
0 0
1
Now Insert 34

38. Double rotation (inside case)
38
3
0
3
20
10 30
25
1
40
2
5
0
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

39. Extended Example
Insert 3,2,1,4,5,6,7, 16,15,14
3
Fig 1
3
2
Fig 2
3
2
1
Fig 3
2
1 3
Fig 4
2
1 3
4
Fig 5
2
1 3
4
5
Fig 6

40. 2
1 4
53
Fig 7 6
2
1 4
53
Fig 8
4
2 5
61 3
Fig 9
4
2 5
61 3
7Fig 10
4
2 6
71 3
5 Fig 11

41. 4
2 6
71 3
5 16
Fig 12
4
2 6
71 3
5 16
15
Fig 13
4
2 6
151 3 5
16
7Fig 14

42. 5
4
2 7
151 3 6
1614
Fig 16
4
2 6
151 3 5
16
7
14
Fig 15
Deletions can be done with similar rotations

43. 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.
43

44. 44
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:
1. Difficult to program & debug; more space for balance factor.
2. Asymptotically faster but rebalancing costs time.
3. Most large searches are done in database systems on disk and use
other structures (e.g. B-trees).
4. 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

45. Deleting a Node from an AVL Tree
• Deletion of a node in an AVL tree is similar to that of binary search trees.
• But deletion may disturb the AVLness of the tree, so to re-balance the
AVL tree we need to perform rotations.
• There are two classes of rotation that can be performed on an AVL tree
after deleting a given node: R rotation and L rotation.
• If the node to be deleted is present in the left sub-tree of the critical
node, then L rotation is applied else if node is in the right sub-tree, R
rotation is performed.
• Further there are three categories of L and R rotations. The variations of
L rotation are: L-1, L0 and L1 rotation. Correspondingly for R rotation,
there are R0, R-1 and R1 rotations.

46. Deleting a Node from an AVL Tree
• R0 Rotation
 Let B be the root of the left or right sub-tree of A (critical node).
 R0 rotation is applied if the balance factor of B is 0.
 Consider the AVL tree given below and delete 72 from it.
4
5
6
3
3
6
2
7 3
9
7
2
-1
1
1
-1
0
1
8
0
4
0
0
0
45
6336
27 39
0
2
1
-1
0
18 40
0 0
36
4527
18
39
1
-1
0
-1
1
40
63
0
0

47. Deleting a Node from an AVL Tree
• R1 Rotation
 Let B be the root of the left or right sub-tree of the critical node.
 R1 rotation is applied if the balance factor of B is 1.
 Consider the AVL tree given below and delete 72 from it.
4
5
6
3
3
6
2
7 3
9
7
2
1
1
1
0
1
1
8
0
0
45
6336
27 39
0
2
1
0
1
18
0
3
6
4
5
2
7
1
8
3
9
0
0
0 0
1
6
3
0

48. Deleting a Node from an AVL Tree
• R-1Rotation
 Let B be the root of the left or right sub-tree of the critical node.
 R-1 rotation is applied if the balance factor of B is -1.
 Consider the AVL tree given below and delete 72 from it.
4
5
6
3
3
6
2
7 3
9
-1
1
0 0
-1
3
7
4
1
0 0
7
2
0
4
5
6
3
3
6
2
7 3
9
0
2
0 0
-1
3
7
4
1
0 0
3
9
4
5
3
6
2
7
3
7
4
1
1
1
1 0
1
0
6
3
0