CSE 326: Data Structures Lecture #9 AVL II Alon Halevy Spring Quarter 2001 Alright, today we’ll get a little Yin and Yang. We saw B-Trees, but they were just too hard to use! Let’s see something easier! (a bit)

Deletion (Really Easy Case) 1 2 3 Delete(17) 10 5 15 2 9 12 20 OK, if we have a bit of extra time, do this. Let’s try deleting. 15 is easy! It has two children, so we do BST deletion. 17 replaces 15. 15 goes away. Did we disturb the tree? NO! 3 17 30

Deletion (Pretty Easy Case) 1 2 3 Delete(15) 10 5 15 2 9 12 20 OK, if we have a bit of extra time, do this. Let’s try deleting. 15 is easy! It has two children, so we do BST deletion. 17 replaces 15. 15 goes away. Did we disturb the tree? NO! 3 17 30

Deletion (Pretty Easy Case cont.) 3 Delete(15) 10 2 2 5 17 1 1 2 9 12 20 OK, if we have a bit of extra time, do this. Let’s try deleting. 15 is easy! It has two children, so we do BST deletion. 17 replaces 15. 15 goes away. Did we disturb the tree? NO! 3 30

Deletion (Hard Case #1) Delete(12) 10 5 17 2 9 12 20 3 30 3 2 1 2 3 Delete(12) 10 5 17 2 9 12 20 Now, let’s delete 12. 12 goes away. Now, there’s trouble. We’ve put an imbalance in. So, we check up from the point of deletion and fix the imbalance at 17. 3 30

Single Rotation on Deletion 1 2 3 3 10 10 2 1 5 17 5 20 1 2 9 20 2 9 17 30 But what happened on the fix? Something very disturbing. What? The subtree’s height changed!! So, the deletion can propagate. 3 30 3 What is different about deletion than insertion?

Deletion (Hard Case) Delete(9) 10 5 17 2 9 12 12 20 20 3 11 15 15 18 3 4 Delete(9) 10 5 17 2 9 12 12 20 20 Now, let’s delete 12. 12 goes away. Now, there’s trouble. We’ve put an imbalance in. So, we check up from the point of deletion and fix the imbalance at 17. 1 1 3 11 15 15 18 30 30 13 13 33 33

Double Rotation on Deletion Not finished! 1 2 3 4 2 1 3 4 10 10 5 17 3 17 2 2 12 20 2 5 12 20 1 1 1 1 3 11 15 18 30 11 15 18 30 13 33 13 33

Deletion with Propagation 2 1 3 4 10 What different about this case? 3 17 2 5 12 20 1 1 We get to choose whether to single or double rotate! 11 15 18 30 13 33

Propagated Single Rotation 2 1 3 4 4 10 17 3 2 3 17 10 20 1 2 1 2 5 12 20 3 12 18 30 1 1 1 11 15 18 30 2 5 11 15 33 13 33 13

Propagated Double Rotation 2 1 3 4 4 10 12 2 3 3 17 10 17 1 1 2 2 5 12 20 3 11 15 20 1 1 1 11 15 18 30 2 5 13 18 30 13 33 33

AVL Deletion Algorithm Recursive If at node, delete it Otherwise recurse to find it in 3. Correct heights a. If imbalance #1, single rotate b. If imbalance #2 (or don’t care), double rotate Iterative 1. Search downward for node, stacking parent nodes 2. Delete node 3. Unwind stack, correcting heights a. If imbalance #1, single rotate b. If imbalance #2 (or don’t care) double rotate OK, here’s the algorithm again. Notice that there’s very little difference between the recursive and iterative. Why do I keep a stack for the iterative version? To go bottom to top. Can’t I go top down? Now, what’s left? Single and double rotate!

Fun with AVL Trees Input: sequence of n keys (unordered) 19 3 4 18 7 19 3 4 18 7 Insert each into initially empty AVL tree Print using inorder traversal O(n) Result? Are we having fun yet?

Is There a Faster Way? But suppose input is already sorted 3 4 7 18 19 3 4 7 18 19 Can we do better than O(n log n)?

AVL buildTree 5 8 10 15 17 20 30 35 40 Divide & Conquer 17 Divide the problem into parts Solve each part recursively Merge the parts into a general solution 17 IT DEPENDS! How long does divide & conquer take? 8 10 15 5 20 30 35 40

BuildTree Example 5 8 10 15 17 20 30 35 40 3 17 5 8 10 15 2 2 20 30 35 40 10 35 20 30 5 8 1 1 8 15 30 40 5 20

BuildTree Analysis (Approximate) T(n) = 2T(n/2) + 1 T(n) = 2(2T(n/4)+1) + 1 T(n) = 4T(n/4) + 2 + 1 T(n) = 4(2T(n/8)+1) + 2 + 1 T(n) = 8T(n/8) + 4 + 2 + 1 T(n) = 2kT(n/2k) + let 2k = n, log n = k T(n) = nT(1) + T(n) = (n) Summation is 2^logn + 2^logn-1 + 2^logn-2+… n+n/2+n/4+n/8+… ~2n

BuildTree Analysis (Exact) Precise Analysis: T(0) = b T(n) = T( ) + T( ) + c By induction on n: T(n) = (b+c)n + b Base case: T(0) = b = (b+c)0 + b Induction step: T(n) = (b+c) + b + (b+c) + b + c = (b+c)n + b QED: T(n) = (b+c)n + b = (n)

Application: Batch Deletion Suppose we are using lazy deletion When there are lots of deleted nodes (n/2), need to flush them all out Batch deletion: Print non-deleted nodes into an array How? Divide & conquer AVL Treebuild Total time:

Thinking About AVL Observations + Worst case height of an AVL tree is about 1.44 log n + Insert, Find, Delete in worst case O(log n) + Only one (single or double) rotation needed on insertion - O(log n) rotations needed on deletion + Compatible with lazy deletion - Height fields must be maintained (or 2-bit balance)

Alternatives to AVL Trees Weight balanced trees keep about the same number of nodes in each subtree not nearly as nice Splay trees “blind” adjusting version of AVL trees no height information maintained! insert/find always rotates node to the root! worst case time is O(n) amortized time for all operations is O(log n) mysterious, but often faster than AVL trees in practice (better low-order terms)