Binary Search Trees

1. Binary Search Trees data compare data data T < < We can search the tree “efficiently” log n < data data ITK 179

2. O(log2n) What is the real benefit of using BST? Search 18, 2, 26 20 10 30 1 17 25 35 0 5 14 19 28 34 38 23 4 7 12 15 18 22 27 32 36 3 11 13 21 26 37 ITK 279

3. What is the benefit of using BST? The average height of a binary search tree of n nodes is O(log2n) if the n nodes are arrived based on a uniform distribution on the space of possible keys. This is as good as the binary search on a sorted array. Proof: not so difficult, not so trivial.. f(n) : The average internal path length of an n-node BST ITK 279

4. The average height of a BST : O(log2n) f(n) : The average internal path length of an n-node BST f(3) : (3+2+3) / 3 = 2.67 0 0 1 1 1 1 2 2 0+1+1=2 0+1+2=3 0+1+2=3 ITK 279

5. Analysis Average Heights on n Random Keys BST O(log2n) How bad is the constant?  log2n + O(log2log2n)  = 4.31101… Devroye and Reed,, SIAM J. Comput. ‘95 ITK 279

6. The sequence of node insertion will affect the shape of the BST 1 Highly unbalanced BST 7 12 1, 7, 12, 25, 27, 13, 23, 17, 15 25 13 27 This situation is not uncommon e.g., the data is roughly sorted. 23 17 15 ITK 179

7. Highly unbalanced BST 1 7 Balanced BST 12 17 25 13 27 13 25 7 15 23 27 23 17 1 12 15 ITK 179

8. AVL Trees(Ch 11.3, 11.4) -- an example of Dynamic trees • Adel’son-Vels’kii and Landis, Soviet Mathematic Doklady, 3:1259-1263, 1962 C. Crane, D. Knuth, et al in 1970’s We dynamicallymaintain the properties of AVL-tree when we insert (remove) a node by four different operations (rotations) Dynamic trees Static trees Huffman Code -- an example of Static trees We staticallyanalyze the code and build up an optimal tree for retrieving the code words. ITK 179

9. AVL Tree A BST in which the height difference between the two children of any node is always less than 2. +1 +1 -1 j k j+1 k+1 h h+1 ITK 179

10. performed at the bad node where the difference between the heights of its two children is bigger than 1 after insertion. Four operations : (rotations) If a node is bad caused by: then perform • Right-child’s Right-child • Left-child’s Left-child • Right-child’s Left-child • Left-Child’s Right-child • RR rotation • LL rotation • RL rotation • LR rotation ITK 179

11. No Rotation is Needed +1 0 -1 0 ITK 179

12. RR Rotations: +2 +1 ITK 179

13. RR Rotation: +2 R +0 R +0 +1 ITK 179

14. LL Rotation: L -2 +0 L -1 +0 ITK 179

15. Rotations: RL +1 +2 0 h+1 +1 -1 0 0 h+1 +0 -1 h h ITK 179

16. Rotations: RL RL Rotation: R 0 +2 L h+1 -1 0 +1 h+1 -1 h h h+1 h+1 h h ITK 179

17. Rotations: LR LR Rotation: L -2 0 1 R h+1 0 +1 h+1 -1 -1 h h h+1 h+1 h h ITK 179

18. Rotate this sub-tree first +2 R Could be RR or RL, depending on what happens in the blue sub-tree. R +2 Afterwards, examine the red node again to see is another rotation is needed. ITK 179

19. Rotations: LL +2 +1 +1 h+1 -1 -2 -1 h -2 -1 0 h-1 h -1 0 h-1 h-1 ITK 179

20. Example: 10 20 30 10 RR Rotation: 20 20 10 30 30 ITK 179

21. Example: 40 50 20 RR Rotation: 20 10 30 10 40 40 50 30 50 ITK 179

22. Example: 35 20 R 30 RL Rotation: 10 40 40 L 20 30 50 10 50 35 35 ITK 179

23. Possible complications Re-assign the links +2 h+1 -1 h+1 -1 h h Tracking the heights and balance-factors ITK 179

24. RR rotation in JAVA a=t +2 R // RR rotation on t; privatevoid RR(TNode<E> t){ TNode<E> a = t, b = t.right; a.right = b.left; b.left = a; a.height -= 2; return b; } b h R +1 h h height is an extra member variable in the TNode<E>. ITK 179

25. RL rotation in C++ a=t // RL rotation on t; privatevoid RL(TNode<E> t){ TNode<E>a, b, c; a = t; b = t.right; c = t.right.left; a.right = c.left; b.left = c.right; c.left = a; c.right = b; c.height++; b.height--; a.height-=2; return c; } b R +2 L c h+1 -1 h+1 -1 h h ITK 179

26. How to determine which rotation is needed? RR L L R L L R +2 -2 +2 -2 -1 +1 +1 -1 ITK 179

27. AVL h: Average Heights n Random Keys ITK 179