This is an PPT of Data Structure. It include the following topic "AVL Tree in Data Structure"

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

23. 5
7

24. 5
7
19

25. 5
7
19
19
7
5

26. 19
7
5

27. 19
7
5
12

28. 19
7
5
12
10

29. 19
7
5
12
10
7
5
19
12
10

30. 7
5
19
12
10

31. 7
5
19
12
10
15

32. 7
5
19
12
10
15
12
7 19
155 10

33. 12
7 19
155 10

34. 12
7 19
155 10
18

35. 12
7 19
155 10
18
12
7
5 10
18
15 19

36. 12
7
5 10
18
15 19

37. 12
7
5 10
18
15 19
20

38. 12
7
5 10
18
15 19
20
25

39. 12
7
5 10
18
15 19
20
25

40. 12
7
5 10
18
15 20
19 25

41. 12
7
5 10
18
15 20
19 25
23

42. 12
7
5 10
18
15 20
19 25
23

43. 12
7
5 10
20
18 25
2315 19

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Data Structure by ABDUL BARI