Fundamental Algorithms
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1 Fundamental Algorithms Chapter 6: AVL Trees Michael Bader Winter 2011/12 Chapter 6: AVL Trees, Winter 2011/12 1
2 Part I AVL Trees Chapter 6: AVL Trees, Winter 2011/12 2
3 Binary Search Trees Summary Complexity of Searching: worst-case complexity depends on height of the search trees O(log n) for balanced trees Inserting and Deleting: insertion and deletion might change balance of trees question: how expensive is re-balancing? Test: Inserting/Deleting into a (fully) balanced tree strict balancing (uniform depth for all leaves) too strict Chapter 6: AVL Trees, Winter 2011/12 3
4 Height Balance Definition (height balance) Let h(x) be the height of a binary tree x. Then, the height balance b(x.key) of a node x.key is defined as b(x.key) := h(x.rightchild) h(x.leftchild) i.e. the difference of the heights of the two subtrees of x.key. Example: Chapter 6: AVL Trees, Winter 2011/12 4
5 AVL Trees Definition (AVL tree) A binary search tree x is called an AVL tree, if: 1. b(x.key) { 1, 0, 1}, and 2. x.leftchild and x.rightchild are both AVL trees. = the height balance of every node must be -1, 0, or 1 Chapter 6: AVL Trees, Winter 2011/12 5
6 Number of Nodes in an AVL Tree Crucial Question for Complexity: What is the maximal and minimal number of nodes that can be stored in an AVL tree of given height h? Maximal number: a full binary tree has a height balance of 0 for every node thus: full binary trees are AVL trees number of nodes: 2 h 1 nodes. Chapter 6: AVL Trees, Winter 2011/12 6
7 Minimal Number of Nodes in an AVL Tree Minimal number: A minimal AVL tree of height h consists of a root node one subtree that is a minimal AVL tree of height h 1 one subtree that is a minimal AVL tree of height h 2 leads to recurrence: N minavl (h) = 1 + N minavl (h 1) + N minavl (h 2) In addition, we know that a minimal AVL tree of height 1 has 1 node: N minavl (1) = 1 a minimal AVL tree of height 2 has 2 nodes: N minavl (2) = 2 Chapter 6: AVL Trees, Winter 2011/12 7
8 Minimal Number of Nodes in an AVL Tree (2) Solve recurrence: N minavl (h) = 1 + N minavl (h 1) + N minavl (h 2) N minavl (1) = 1 N minavl (2) = 2 Compare with Fibonacci numbers: f n = f n 1 + f n 2, f 0 = f 1 = 1 h f h N minavl (h) Claim: N minavl (h) = f h+1 1 (proof by induction) Chapter 6: AVL Trees, Winter 2011/12 8
9 Minimal Number of Nodes in an AVL Tree (2) Minimum number of nodes: N minavl (h) = f h+1 1 inequality for Fibonacci numbers: 2 n/2 f n 2 n 2 (h+1)/2 1 N minavl (h) 2 h+1 1 thus: an AVL tree that contains n nodes will be of height Θ(log n) Corollaries: Searching in an AVL tree has a time complexity of Θ(log n) Inserting, or deleting a single element in an AVL tree has a time complexity of Θ(log n) BUT: standard inserting/deleting will probably destroy the AVL property. Chapter 6: AVL Trees, Winter 2011/12 9
10 Part II Algorithms on AVL Trees Chapter 6: AVL Trees, Winter 2011/12 10
11 Inserting and Deleting on AVL Trees General Concept: insert/delete via standard algorithms after insert/delete: load balance b(node) might be changed to +2 or 2 for certain nodes re-balance load after each step Requirements: re-balancing must have O(log n) worst-case complexity Solution: apply certain rotation operations Chapter 6: AVL Trees, Winter 2011/12 11
12 Left Rotation Situation: height balance of the node is +2 (or larger), and height balance of the right subtree is 0, or +1 can reduce height of the subtree and require further rotations Chapter 6: AVL Trees, Winter 2011/12 12
13 Left Rotation Implementation LeftRotAVL ( x : BinTree ) { } x := ( x. r i g h t C h i l d. key, ( x. key, x. l e f t C h i l d, x. r i g h t C h i l d. l e f t C h i l d ), x. r i g h t C h i l d. r i g h t C h i l d ) ; Chapter 6: AVL Trees, Winter 2011/12 13
14 Right Rotation Situation: height balance of the node is -2 (or smaller), and height balance of the left subtree is 0, or -1 can reduce height of the subtree and require further rotations Chapter 6: AVL Trees, Winter 2011/12 14
15 Right Rotation Implementation RightRotAVL ( x : BinTree ) { } x := ( x. l e f t C h i l d. key, x. l e f t C h i l d. l e f t C h i l d, ( x. key, x. l e f t C h i l d. r i g h t C h i l d, x. r i g h t C h i l d ) ) ; Chapter 6: AVL Trees, Winter 2011/12 15
16 Right-Left Rotation Situation: height balance of the node is +2, and height balance of the right subtree is -1 left rotation successful? left rotation is not sufficient to restore the AVL property Chapter 6: AVL Trees, Winter 2011/12 16
17 Solution: Two Successive Rotations Chapter 6: AVL Trees, Winter 2011/12 17
18 Right-Left Rotation Implementation RightRotLeftRot ( x : BinTree ) { RightRotAVL ( x. r i g h t C h i l d ) ; LeftRotAVL ( x ) ; } In a single procedure: RightLeftRotAVL ( x : BinTree ) { x := ( x. r i g h t C h i l d. l e f t C h i l d. key, ( x. key, x. l e f t C h i l d, x. r i g h t C h i l d. l e f t C h i l d. l e f t C h i l d ), ( x. r i g h t C h i l d. key, x. r i g h t C h i l d. l e f t C h i l d. r i g h t C h i l d, x. r i g h t C h i l d. r i g h t C h i l d ) ) ; } Chapter 6: AVL Trees, Winter 2011/12 18
19 Left-Right Rotation Situation: height balance of the node is -2, and height balance of the right subtree is +1 Analogous to Right-Left Rotation: again, right rotation is not sufficient to restore the AVL property right rotation followed by a left rotation Chapter 6: AVL Trees, Winter 2011/12 19
20 Left-Right Rotation Scheme Chapter 6: AVL Trees, Winter 2011/12 20
21 Left-Right Rotation Implementation LeftRotRightRot ( x : BinTree ) { LeftRotAVL ( x. l e f t C h i l d ) ; RightRotAVL ( x ) ; } In a single procedure: LeftRightRotAVL ( x : BinTree ) { x := ( x. l e f t C h i l d. r i g h t C h i l d. key, ( x. l e f t C h i l d. key, x. l e f t C h i l d. l e f t C h i l d, x. l e f t C h i l d. r i g h t C h i l d. l e f t C h i l d ), ( x. key, x. l e f t C h i l d. r i g h t C h i l d. r i g h t C h i l d, x. r i g h t C h i l d ) ) ; } Chapter 6: AVL Trees, Winter 2011/12 21
22 Effect of Rotation Operations Observations: LeftRightRot and RightLeftRot always reduce the height of the (sub-)tree by 1. LeftRotAVL and RightRotAVL reduce the height by at most 1 (might keep it unchanged) thus: AVL property can be violated in a parent node! After inserting one node into an AVL tree: at most one rotation required to restore AVL property Chapter 6: AVL Trees, Winter 2011/12 22
23 Effect of Rotation Operations (2) After inserting one node into an AVL tree: at most one rotation required to restore AVL property After deleting one node of an AVL tree: up to log n rotations can be required to restore AVL property For insertion and deletion: imbalance might occur on a much higher level thus: AVL property has to checked in the entire branch of the tree (up to the root) Corollary: Time complexity for deleting, inserting, and searching in an AVL tree is O(log n) Chapter 6: AVL Trees, Winter 2011/12 23
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