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![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)](https://image.slidesharecdn.com/14akashkumarchaubey-200413102920/75/AVL-Tree-in-Data-Structure-5-2048.jpg)
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AVL Tree in Data Structure
- 1. PRESENTED BY- AKASH KUMARCHAUBEY CLASS – MCA(1) SEMESTER : 2 ROLL NO. –MCA/25014/18 AVL TREE
- 2. WHAT IS BINARYSEARCHTREE • 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 disadvantageofskewed 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 isa 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 treeis 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.
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- 8. If afterthe 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 therotation 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 rotationswitches 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 nodeis 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.
- 13. Inserted nodeis 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
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- 15. Single rotationdoesnot 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 nodeis 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
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- 18. Inserted nodeis 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.
- 20. // Left LeftCase 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); }
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