IKI 10100: Data Structures & Algorithms Ruli Manurung (acknowledgments to Denny & Ade Azurat) 1 Fasilkom UI Ruli Manurung (Fasilkom UI)IKI10100: Lecture27 th Mar 2007 AVL Tree

2 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 AVL Tree Definition Properties Operations Outline

3 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 XX AVL Trees Unbalanced Binary Search Trees are bad. Worst case: operations take O(n). AVL (Adelson-Velskii & Landis) trees maintain balance. For each node in tree, height of left subtree and height of right subtree differ by a maximum of 1. H H-1 H-2

4 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 AVL Trees

5 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 AVL Trees

6 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar Insertion for AVL Tree After insert 1

7 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Insertion for AVL Tree To ensure balance condition for AVL-tree, after insertion of a new node, we back up the path from the inserted node to root and check the balance condition for each node. If after insertion, the balance condition does not hold in a certain node, we do one of the following rotations: Single rotation Double rotation

8 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Insertions Causing Imbalance R P Q k1k1 k2k2 Q k2k2 P k1k1 R An insertion into the subtree: P (outside) - case 1 Q (inside) - case 2 An insertion into the subtree: Q (inside) - case 3 R (outside) - case 4 H P =H Q =H R

9 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 A k2k2 B k1k1 C CB A k1k1 k2k2 Single Rotation (case 1) H A =H B +1 H B =H C

10 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Single Rotation (case 4) C k1k1 B k2k2 A AB C k2k2 k1k1 H A =H B H C =H B +1

11 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Q k2k2 P k1k1 R R P Q k1k1 k2k2 Problem with Single Rotation Single rotation does not work for case 2 and 3 (inside case) H Q =H P +1 H P =H R

12 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 C k3k3 A k1k1 D B k2k2 Double Rotation: Step C k3k3 A k1k1 D B k2k2 H A =H B =H C =H D

13 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 C k3k3 A k1k1 D B k2k2 Double Rotation: Step

14 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 C k3k3 A k1k1 D B k2k2 C k3k3 A k1k1 D B k2k2 Double Rotation H A =H B =H C =H D

15 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 B k1k1 D k3k3 A C k2k2 B k1k1 D k3k3 A C k2k2 Double Rotation H A =H B =H C =H D

16 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar Example Insert 3 into the AVL tree

17 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Example Insert 5 into the AVL tree

18 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 AVL Trees: Exercise Insertion order: 10, 85, 15, 70, 20, 60, 30, 50, 65, 80, 90, 40, 5, 55

19 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Remove Operation in AVL Tree Removing a node from an AVL Tree is the same as removing from a binary search tree. However, it may unbalance the tree. Similar to insertion, starting from the removed node we check all the nodes in the path up to the root for the first unbalance node. Use the appropriate single or double rotation to balance the tree. May need to continue searching for unbalanced nodes all the way to the root.

20 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Deletion X in AVL Trees Deletion: Case 1: if X is a leaf, delete X Case 2: if X has 1 child, use it to replace X Case 3: if X has 2 children, replace X with its inorder predecessor (and recursively delete it) Rebalancing

21 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 55 (case 1)

22 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 55 (case 1)

23 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 50 (case 2)

24 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 50 (case 2)

25 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 60 (case 3) prev

26 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 60 (case 3)

27 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 55 (case 3) prev

28 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 55 (case 3)

29 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 50 (case 3) prev

30 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 50 (case 3)

31 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 40 (case 3) prev

32 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 40 : Rebalancing Case ?

33 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Delete 40: after rebalancing Single rotation is preferred!

34 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Minimum Element in AVL Tree An AVL Tree of height H has at least F H+3 -1 nodes, where F i is the i-th fibonacci number S 0 = 1 S 1 = 2 S H = S H-1 + S H S H-1 S H-2 H H-1 H-2

35 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 AVL Tree: analysis (1)

36 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 AVL Tree: analysis (2) The depth of AVL Trees is at most logarithmic. So, all of the operations on AVL trees are also logarithmic. The worst-case height is at most 44 percent more than the minimum possible for binary trees.

37 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar 2007 Summary Find element, insert element, and remove element operations all have complexity O(log n) for worst case Insert operation: top-down insertion and bottom up balancing

38 Ruli Manurung (Fasilkom UI)IKI10100: Lecture 27 th Mar /resources/animation/data- structure/avl/avltree.html Section 19.4 of Weiss book Further Study