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- 1. AVL Trees 6 v 3 8 z 4 AVL Trees 1
- 2. AVL Tree DefinitionAVL trees are 4balanced. 44 2An AVL Tree is a 3 17 78binary search tree 1 2 1such that for every 32 50 88 1internal node v of T, 48 62 1the heights of thechildren of v can differby at most 1. An example of an AVL tree where the heights are shown next to the nodes: AVL Trees 2
- 3. AVL Trees: AVL trees were invented in 1962 by two Russian scientist G.M Adelson Velsky & E.M Landis. Definition: An AVL tree is a binary search tree in which the balancefactor of every node, which is defined as the difference b/w theheights of the node’s left & right sub trees is either 0 or +1or -1 . Balance factor = ht of left sub tree – ht of right sub tree.ht=height
- 4. Example : AVL tree BST ( not AVL tree ) 1 2 10 10 0 5 1 20 05 020 1 4 -1 0 1 -1 7 4 7 12 0 0 0 0 2 8 2 8
- 5. If an insertion of a new node makes an AVL treeunbalanced , we transform the tree by a rotation Rotation : A Rotation in an AVL tree is a local transformation ofits sub tree rooted at a node whose balance has become either+2 or -2 ,if there are several such nodes, we rotate the treerooted at the unbalanced node that is the closest to the newlyinserted leaf.
- 6. Types of rotations : Totally there are four types of rotations1. Single right rotation or R-rotation2. Single left rotation or L-rotation3. Double right-left rotation or RL-rotation4. Double left-right rotation or LR-rotation
- 7. Points to remember to select different rotation technique :1.Straight line with positive unbalanced. apply Right rotation for unbalanced node. +2 5 04 +1 4 0 0 2 5 0 2
- 8. 2.Straight line with negative unbalanced. apply left rotation for unbalanced node. -2 5 06 6 -1 0 0 5 7 70
- 9. 3.Curved line with positive unbalanced. apply left-right rotation. Right rotation for unbalanced node and left rotationfor the nearest node. +2 5 0 4 4 -1 0 2 5 0 2 0
- 10. 4.Curved line with negative unbalanced. apply right-left rotation. Left rotation for unbalanced node and right rotationfor the nearest node. -2 5 0 7 +1 8 0 5 8 0 0 7
- 11. n(2) 3 n(1)Height of an AVL Tree 4 Fact: The height of an AVL tree storing n keys is O(log n). Proof: Let us bound n(h): the minimum number of internal nodes of an AVL tree of height h. We easily see that n(1) = 1 and n(2) = 2 For n > 2, an AVL tree of height h contains the root node, one AVL subtree of height n-1 and another of height n-2. That is, n(h) = 1 + n(h-1) + n(h-2) Knowing n(h-1) > n(h-2), we get n(h) > 2n(h-2). So n(h) > 2n(h-2), n(h) > 4n(h-4), n(h) > 8n(n-6), … (by induction), n(h) > 2in(h-2i) Solving the base case we get: n(h) > 2 h/2-1 Taking logarithms: h < 2log n(h) +2 Thus the height of an AVL Trees is O(log n) AVL tree 11
- 12. Insertion in an AVL Tree Insertion is as in a binary search tree Always done by expanding an external node. Example: 44 44 c=z 17 78 17 78 a=y 32 50 88 32 50 88 48 62 48 62 b=x 54 w before insertion after insertion AVL Trees 12
- 13. Trinode Restructuring let (a,b,c) be an inorder listing of x, y, z perform the rotations needed to make b the topmost node of the three (other two cases a=z case 2: double rotation a=z are symmetrical) (a right rotation about c, c=y then a left rotation about a) b=y T0 T0 c=x b=x T3 T1 b=y b=x T1 T2 T2 T3 a=z c=x a=z c=ycase 1: single rotation(a left rotation about a) T0 T1 T2 T3 T0 T1 T2 T3 AVL Trees 13
- 14. Insertion Example, continued 5 44 2 z 64 17 78 7 3 2y 1 1 32 1 50 4 88 1 2 x 48 1 3 62 5 54 T3unbalanced... T2 T0 4 T1 44 4 2 3 x 17 2 y 62 z6 1 2 2 32 1 1 50 3 5 78 7 1 1 ...balanced 48 54 88 T2 T0 T1 T3 AVL Trees 14
- 15. Restructuring(as Single Rotations) Single Rotations: a=z single rotation b=y b=y a=z c=x c=x T0 T3 T1 T3 T0 T1 T2 T2 c=z single rotation b=y b=y a=x c=z a=x T3 T3 T0 T2 T2 T1 T0 T1 AVL Trees 15
- 16. Restructuring(as Double Rotations) double rotations: a=z double rotation b=x c=y a=z c=y b=x T0 T2 T2 T3 T0 T1 T3 T1 c=z double rotation b=x a=y a=y c=z b=x T0 T2 T3 T2 T3 T1 T0 T1 AVL Trees 16
- 17. Removal in an AVL Tree Removal begins as in a binary search tree, which means the node removed will become an empty external node. Its parent, w, may cause an imbalance. Example: 44 44 17 62 17 62 32 50 78 50 78 48 54 88 48 54 88 before deletion of 32 after deletion AVL Trees 17
- 18. Rebalancing after a Removal Let z be the first unbalanced node encountered while travelling up the tree from w. Also, let y be the child of z with the larger height, and let x be the child of y with the larger height. We perform restructure(x) to restore balance at z. As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached 62 a=z 44 44 78 w 17 62 b=y 17 50 88 50 78 c=x 48 54 48 54 88 AVL Trees 18
- 19. Running Times forAVL Trees a single restructure is O(1) using a linked-structure binary tree find is O(log n) height of tree is O(log n), no restructures needed insert is O(log n) initial find is O(log n) Restructuring up the tree, maintaining heights is O(log n) remove is O(log n) initial find is O(log n) Restructuring up the tree, maintaining heights is O(log n) AVL Trees 19

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