1. Oct 26, 2001CSE 373, Autumn 20011 A Forest of Trees Binary search trees: simple. –good on average: O(log n) –bad in the worst case: O(n) AVL trees: more complicated. –good worst case bound: O(log n) Splay trees: “medium” complexity of code. –guaranteed good performance over a long sequence of operations: O(m log n) over m operations (amortized running time).

2. Oct 26, 2001CSE 373, Autumn 20012 Self-adjusting structures To avoid repeated accesses to deep nodes, the (tree) structure needs to alter itself after each operation. Idea: Since AVL rotation seems to reduce the depth of nodes, do an AVL rotate (from node to root) any time we do an operation.

3. Oct 26, 2001CSE 373, Autumn 20013 Idea does not work 5 4 6 1 7 3 2 find(1) 5 4 6 1 7 3 2 find(2), find(3), find(4), …

4. Oct 26, 2001CSE 373, Autumn 20014 Splay operation Splay(K): Search for K in the usual way. Let X be the last node inspected. –If K is in the tree, then K is in X. –If K is not in the tree, then P has an empty child where the search for K terminated. Follow the path from X to the root, carrying out rotations along the way.

5. Oct 26, 2001CSE 373, Autumn 20015 Case 0: X is the root Do nothing! X A

6. Oct 26, 2001CSE 373, Autumn 20016 Case 1: no grandparent X has no grandparent (i.e., X’s parent is the root). Perform a single rotation on X and X’s parent. P C B X A X C B P A

7. Oct 26, 2001CSE 373, Autumn 20017 Case 2: zig-zig X and X’s parent are both left children or both right children. Perform two single rotations: –X’s grandparent and X’s parent. –X and X’s parent. P C B X A G D P C B G A X D

8. Oct 26, 2001CSE 373, Autumn 20018 Case 3: zig-zag One of X and X’s parent is a left child and the other is a right child. Perform a double rotation. P C B X A G D P C B G A X D

9. Oct 26, 2001CSE 373, Autumn 20019 Properties of Splay(K) If K is in the tree, then the resulting BST will have K at the root. If K is not in the tree, then the root contains a key that would be the successor or predecessor of K, if K were in the tree. (What determines whether the successor predecessor is at the root?)

10. Oct 26, 2001CSE 373, Autumn 200110 Splay trees: find, insert Find(K): Do Splay(K). Examine the root to see if it has K. Insert(I, K): Do Splay(K). If K is at the root, replace its element with I. Otherwise, create a new node from K and I and break one link to make this node the new root. T2T2 J T1T1 K T2T2 J T1T1 T

11. Oct 26, 2001CSE 373, Autumn 200111 Splay tree: remove Remove(K): Do Splay(K). If the root does not contain K, do nothing. Otherwise, delete the root and Concat the two subtrees of the root. T2T2 K T1T1 T T2T2 T1T1 Splay(findMax(T 1 )) T2T2 M