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Lecture 12

Lecture 12. AVL Trees. Prof. Sin-Min Lee Department of Computer Science. Types of Trees. binary search trees. AVL trees. 2-3 trees. tries. trees. dynamic. static. game trees. search trees. priority queues and heaps. graphs. Huffman coding tree. Complete Binary Tree.

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Lecture 12

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  1. Lecture 12 AVL Trees Prof. Sin-Min Lee Department of Computer Science

  2. Types of Trees binary search trees AVL trees 2-3 trees tries trees dynamic static game trees search trees priority queues and heaps graphs Huffman coding tree

  3. Complete Binary Tree • Is a tree with height h where: • Every node is full upto level h-2, • Level h-1 is completely filled. • Level h is filled from left to right. • Yields to array representation. How to compute indices of parent and child? • How many internal nodes and leafs?

  4. AVL Trees • Balanced binary search tree offer a O(log n) insert and delete. • But balancing itself costs O(n) in the average case. • In this case, even though delete will be O(log n), insert will be O(n). • Is there any way to have a O(log n) insert too? • Yes, by almost but not fully balancing the tree : AVL (Adelson Velskii and Landis) balancing

  5. Height of a Tree • Definition is same as level. Height of a tree is the length of the longest path from root to some leaf node. • Height of a empty tree is -1. • Height of a single node tree is 0. • Recursive definition: • height(t) = 0 if number of nodes = 1 = -1 if T is empty = 1+ max(height(LT), height(RT)) otherwise

  6. AVL Property • If N is a node in a binary tree, node N has AVL property if the heights of the left sub-tree and right sub-tree are equal or if they differ by 1. • Lets look at some examples.

  7. AVL Tree: Example

  8. Non-AVL Tree

  9. Transforming into AVL Tree • Four different transformations are available called : rotations • Rotations: single right, single left, double right, double left • There is a close relationship between rotations and associative law of algebra.

  10. Transformations • Single right : ((T1 + T2) + T3) = (T1 + (T2 + T3) • Single left : (T1 + (T2 + T3)) = ((T1 + T2) + T3) • Double right : ((T1 + (T2 + T3)) + T4) = ((T1+T2) + (T3+T4)) • Double left : (T1 ((T2+T3) +T4)) = ((T1+T2) + (T3+T4))

  11. Example: AVL Tree for Airports • Consider inserting sequentially: ORY, JFK, BRU, DUS, ZRX, MEX, ORD, NRT, ARN, GLA, GCM • Build a binary-search tree • Build a AVL tree.

  12. Binary Search Tree for Airport Names ORY ZRH JFK MEX BRU ORD DUS ARN NRT GLA GCM

  13. X1 X2 X2 X3 X1 X3 X2 X3 AVL Balancing : Four Rotations X1 X3 Double right X2 X2 X1 Single right X3 X1 X1 X3 X2 Single left Double left X2 X1 X3 X1 X2 X3

  14. ORY ORY JFK JFK BRU BRU An AVL Tree for Airport Names After insertion of ORY, JFK and BRU : Single right Not AVL balanced AVL Balanced

  15. ORY JFK JFK BRU BRU ORY DUS MEX ZRH ORD NRT An AVL Tree for Airport Names (contd.) After insertion of DUS, ZRH, MEX and ORD After insertion of NRT? DUS MEX ZRH ORD Still AVL Balanced

  16. JFK JFK Double Left ORY BRU BRU ORY DUS NRT ZRH DUS MEX ZRH MEX ORD ORD NRT An AVL Tree … Not AVL Balanaced Now add ARN and GLA; no need for rotations; Then add GCM

  17. JFK ORY ORY DUS ARN BRU BRU NRT NRT ZRH ZRH GLA MEX ORD MEX ORD GCM An AVL Tree… JFK GCM ARN DUS GLA Double left NOT AVL BALANCED

  18. For successful search, average number of comparisons: sum of all (path length+1) / number of nodes For the binary search tree (of airports) it is: 39/11 = 3.55 For the AVL tree (of airports) it is : 33/11 = 3.0 Search Operation

  19. Known Performance Results of AVL trees • AVL tree is a sub optimal solution. • How to evaluate its performance? • Bounds (upper bound and lower bound) for number of comparisons: C > log(n+1) + 1 C < 1.44 log(n+2) - 0.328 • AVL trees are no more than 44% worse than optimal trees.

  20. Proposed by Adel’son-Vel’s Skii G.M. and Landis Y.M. (1962) The definition of an AVL tree indicates that the minimum number of nodes in a tree is determined by the recurrence equation: AVLh= AVLh-1 + AVLh-2 +1 where AVL0 = 0 and AVL1 =1. Example: If the height is 1, there must be at least 2 nodes in the AVL tree.

  21. BALANCE FACTORS: • • Indicated by the numbers in the nodes • • Equal the height of right subtree minus the height of left subtree • The height of empty node in AVL is -1; otherwise it is 0 • • All numbers should be +1, 0, or -1 EXAMPLES OF AVL TREES: +1 +1 -1 -1 -1 0 -1 0 0 +1 0 0 0

  22. 1) FIRST CASE: -Resulted from inserting a node in the right subtree of the right child. Q P 0 +2 P +1 Q P +1 +1 0 Q 0 +1 New node inserted AVL tree is unbalanced. The AVL tree is now balanced

  23. 2) SECOND CASE: -Resulted from inserting a node into the left subtree of the right child. (it is more complex.) P P +2 +1 +1 P Q Q -1 Q -1 0 A +1 -1 R A R R R P 0 +2 Q P R +1 0 +2 Q 0 A

  24. FACTS ABOUT NODE DELETION ANDINSERTION: •Require at most 1.44lg(n+2) searches. • Require 1 single or 1 double rotation, depend the balance factors • Deletion can require 1.44lg(n+2) rotations in the worst case. • May or may not need rotation at all

  25. • If append a node to a path of all zero balance factor in AVL tree , no rotations required except updating the the balance factors of nodes. 0 -1 0 +1 0 -1 0 +1 Before updating the nodes After updating the nodes

  26. Example of insertion (with rotation): 20 40 60 . 20 20 40 40 40 20 60 60

  27. Example of insertion (with rotation): 20 40 60 10 15 40 40 20 20 20 60 40 40 20 60 10 60 15 40 15 60 10 20 20

  28. Example of insertion (with rotation) 20 40 60 10 1512 . 40 40 20 20 20 60 40 40 20 60 10 60 15 40 40 15 40 15 60 15 60 10 10 20 10 20 12 20 60 12 15 15 10 40 11 40 12 20 60 10 12 20 60 11

  29. • Deletion of a node may or may not requires rotation . B +2 A 0 +1 C B A +1 +1 0 0 C Rotation required. 0 0 +1 No rotation required

  30. FACTS ABOUT NODE DELETION ANDINSERTION: (cont.) •22% cases of deletion require rebalancing • 47% cases of insertion require rebalancing • Because the more time-consuming deletion occurs less frequently than insertion, it will not endangering the efficiency of rebalancing AVL trees.

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