The document discusses AVL trees, which are self-balancing binary search trees. It defines AVL trees as binary search trees where the heights of the left and right subtrees of every node differ by at most one. Rotations are used to rebalance the tree after insertions or deletions to maintain this height balance property. There are four types of rotations - left, right, left-right, and right-left - to handle different cases of imbalance. The document provides examples of inserting nodes into AVL trees and the resulting rotations required to maintain balance.

1. PRESENTED BY-
AKASH KUMAR CHAUBEY
CLASS – MCA(1)
SEMESTER : 2
ROLL NO. –MCA/25014/18
AVL TREE

2. WHAT IS BINARY SEARCHTREE
• Every node contain only two
child.
• Small value at left child node
and large value at right child
nodeas compare to Root node.
• The main advantage of BST is
that it is easy to perform
like inserting,
traversing and
operations
searching,
deleting.

3.  The disadvantageof skewed binary search tree is that
theworstcase timecomplexityof a search is O(n).
50
40 60
30 45 55 70
Symmetric
40
50
30
20
10
Skewed

4.  There is a need to maintain the binary search tree to
be of the balanced height, so that it is possible to
obtained for the search option a time complexity of
O(log N) in the worst case.
 One of the most popular balanced tree was introduced
by

5. A binary tree is said to be an AVL tree if T is
a root of tree and T(L) is its left sub tree
and T(R) is its right sub-tree of tree T and
H(T(L)) and H(T(R)) are the heights of the
left and
right sub-trees of T respectively, and
|H(T(L))
- H(T(R))|<= 1 Then we called T is AVLtree.
Height of left sub-tree minus height of
Right leftsub-tree
[H(T(L)) - H(T(R))]
Note:
 An emptybinary tree isan AVLTree
 Balance Factorshouldbe
-1 0 1
T
T(L) T(R)

6. T
T(L) T(R)

7. 40
50
30

8.  If after the insertion of the element the, the
balance factor of any node is affect this
problem is overcome by using rotation.
 Rotation is use to restore the balance of
search tree.

9.  Toperform the rotation
it is necessary toidentify
a specific node A whose
balance factor (BF) is
neither 0, -1 0r 1 and
which is the nearest
ancestor to the inserted
node on the path from
the inserted node to the
root.
40
50
30

10.  Single rotation switches the roles of the parent and
child while maintaining the searchorder.
 Werotatea nodeand itschild, child becomes parent
 Parent becomes Right child in LLRotation.
 Parent becomes Left child in RR Rotation.

11.  Inserted node is in the
left sub-tree of the left
sub-tree of A.
 For LL Rotations identify
A and B, where B is a
Left child of A because
insertion is on leftside.
 Then make A as a child
of B.
40
50
30

12. 40
50
30
50
40
30
Parentbecomes
Rightchild
50
40
30

13.  Inserted node is in the
right sub-tree of the
Right sub-tree of A.
 For RR Rotations
identify A and B,
where B is a Right
child of A because
insertion is on Right
side.
 Then make A as a child
of B.
60
50
70

14. 60
50
70
70
60
50
Parentbecomes
Left child
50
60
70

15.  Single rotationdoes not fix the LR rotationand RL
rotation.
 LR and RL rotationsrequireadoublerotation,
involving three node.
 Doublerotation is equivalent toa sequenceof two
single rotation.
 1st rotation on original tree
2nd rotation on the newtree

16.  Inserted node is inthe
right sub-tree of the left
sub-tree of A.
 For LR Rotations identify A
B and C, where B is a Left
child of A and C is right
child of B because Inserted
node is in the right sub-
tree of the left sub-tree of
A.
 Then make A and B as a
child of C.
40
50
45

17. 40
50
45
50
45
40

18.  Inserted node is in theleft
sub-tree of the Right sub-
tree of A .
 For RL Rotations identifyA
B and C, where B is aRight
child of A and C is Left
child of B because Inserted
node is in the left sub-tree
of the Right sub-tree of A.
 Then make A and B as a
child of C.
60
50
55

19. 60
50
55
60
55
50
50
60
55

20. // Left Left Case
if (balance > 1 && key < node->left->key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node->right->key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node->left->key)
{
node->left = leftRotate(node->left);
return rightRotate(node);
}
 // If this node becomes unbalanced, then there are 4 cases





















// Right Left Case
if (balance < -1 && key < node-
>right->key)
{
node->right = rightRotate(node-
>right); return leftRotate(node);
}

22. 5

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24. 5
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25. 5
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26. 19
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27. 19
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28. 19
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29. 19
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30. 7
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31. 7
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32. 7
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33. 12
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34. 12
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35. 12
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36. 12
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37. 12
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38. 12
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39. 12
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42. 12
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43. 12
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44. www.tutorialspoint.com
www.geeksforgeeks.com
Data Structure by ABDUL BARI