AVL Trees / Slide 2. AVL Trees / Slide 4. Let N h be the minimum number of nodes in an AVL tree of height h. AVL Trees / Slide 6

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1 COMP11 Spring 008 AVL Trees / Slide Balanced Binary Search Tree AVL-Trees Worst case height of binary search tree: N-1 Insertion, deletion can be O(N) in the worst case We want a binary search tree with small height. Height of a binary tree with N nodes is at least floor(log N). Goal: To keep the height of a binary search tree O(log N). Balanced binary search trees Examples: AVL trees AVL Trees / Slide AVL Tree An AVL tree is a binary search tree in which Adelson-Velskii and Landis tree is a binary search tree with a balanced condition, in which for every node in the tree, the height of the left and right subtrees differ by at most 1. AVL Trees / Slide AVL Tree with Minimum Number of Nodes AVL tree The AVL property is violated here N 0 = 1 N 1 = N = N = N 1 +N +1= Exercise: find all node heights. Let N h be the minimum number of nodes in an AVL tree of height h AVL Trees / Slide AVL Trees / Slide Height of an AVL Tree Denote N h the minimum number of nodes in an AVL tree of height h. N 0 =1, N 1 = (base) N h = N h-1 + N h- + 1 (recursive relation) & N h-1 > N h- N h+1 > N > N h = N h-1 + N h- + 1 N h >N h- >N h- >8N h- > > i N h-i (by repeated substitution) Smallest AVL tree of height Smallest AVL tree of height 8 Smallest AVL tree of height 9 If h is even, let i=h/ 1 for h-i=. The equation becomes N> h/-1 N N> h/-1 x h=o(logn). If h is odd, let i=(h-1)/ for h-i=1. The equation becomes N> (h-1)/ N 1 N> (h-1)/ x h=o(logn). Thus, many operations (i.e. searching) on an AVL tree will take O(log N) time.

2 AVL Trees / Slide Insertion in AVL Trees Basically follows insertion strategy of binary search tree. But may cause violation of AVL tree property. Restore the destroyed balance condition if needed. AVL Trees / Slide 8 Some Observations After an insertion, only nodes that are on the path from the insertion point to the root might have their balance altered. Because only those nodes have their subtrees altered. Rebalancing the tree at the deepest such node guarantees that the entire tree satisfies the AVL property. 8 8 Original AVL tree Insert Property violated Restore AVL property Node,8, might have balance altered Rebalancing node guarantees the whole tree be AVL AVL Trees / Slide 9 Different Cases for Rebalancing Denote the node that must be rebalanced α (α represents the unbalanced node). Case 1: an insertion into the left subtree of the left child of α (left-left, outside ). Case : an insertion into the right subtree of the left child of α (left-right, inside ). Case : an insertion into the left subtree of the right child of α (right-left, inside ). Case : an insertion into the right subtree of the right child of α (right-right, outside ). Cases 1 & are mirror image symmetries with respect to α, as are cases &. AVL Trees / Slide 10 Rebalance of AVL tree are done with simple modification to tree, known as rotation. Insertion occurs on the outside (i.e., left-left or rightright) is fixed by single rotation of the tree. Left-left means left subtree of the left child of the unbalanced node. Insertion occurs on the inside (i.e., left-right or rightleft) is fixed by double rotation of the tree. Left-right means right subtree of the left child of the unbalance node. AVL Trees / Slide 11 Insertion Algorithm First, insert the new key as a new leaf just as in ordinary binary search tree. Then trace the path from the new leaf towards the root. For each node x encountered, check if heights of left(x) and right(x) differ by at most 1. If yes, proceed to parent(x). If not, re-structure by performing either a single rotation or a double rotation. Note: once we perform a rotation at a node x, we won t need to perform any rotation at any ancestor of x. Rotation (either single or double) needs to perform only once. AVL Trees / Slide 1 Single Rotation to Fix Case 1 k violates An insertion in subtree X, AVL property violated at node k Solution: single rotation AVL-property quiz: Can Y have the same height as the new X? Can Y have the same height as Z?

3 AVL Trees / Slide 1 Single Rotation Case 1 Example AVL Trees / Slide 1 Single Rotation to Fix Case violates k X k An insertion in subtree Z X Case is a symmetric case to case 1. Insertion takes O(Height of AVL Tree) time, takes O(1) time. AVL Trees / Slide Single Rotation Example AVL Trees / Slide If we continue to insert,,, 1, 1, 1, 11, 10, 8, 9 Sequentially insert,, 1,,, to an AVL Tree Insert, 1 Insert 1 at node Insert Insert, at node 1 Insert, at node 1 1 Insert, at node Insert, fine. Insert at node But. Violation remains AVL Trees / Slide 1 Single Rotation Fails to fix Cases & AVL Trees / Slide 18 Double Rotation to Fix Case (left-right) Case : at k because of the insertion in the subtree Y The result after performing the single rotation fails to fix cases & Take case as an example (case is a symmetry to it ). The problem is subtree Y is too deep. doesn t make it less deep. Double rotation to fix case Facts The new key is inserted in the subtree B or subtree C. The AVL-property is violated at k. k -k 1 -k forms a zig-zag shape (left-right). Solution We cannot leave k as the root. The only alternative is to place k as the new root.

4 AVL Trees / Slide 19 Double Rotation to fix Case AVL Trees / Slide 0 Restart our example We ve inserted,, 1,,,,, We ll insert, 1, 1, 1, 11, 10, 8, 9 Facts Double rotation to fix case The new key is inserted in the subtree B or subtree C. The AVL-property is violated at k 1. k -k -k forms a zig-zag shape. Case is a symmetric case to case. k Insert, fine. k Insert at node Double rotation k k AVL Trees / Slide 1 A Insert 1 X k 1 Insert 1 Y k 1 k C 1 Z 1 D 1 1 Double rotation 1 k k 1 AVL Trees / Slide 1 1 Insert Insert AVL Trees / Slide Insert Insert 8, fine then insert 9 (left-right) double rotation (left-right) AVL Trees / Slide More about When the AVL property is lost we can rebalance the tree via rotations. Single Right Rotation (SRR, left-left) Performed when A is unbalanced to the left (the left subtree is higher than the right subtree) and B is leftheavy (the left subtree T1 of B is 1 higher than the right subtree T of B). T1 A SRR at A B T T B T1 A T T

5 AVL Trees / Slide Single Left Rotation (SLR, right-right) performed when A is unbalanced to the right (the right subtree is higher than the left subtree) and B is right-heavy (the right subtree T of B is 1 higher than the left subtree T of B). AVL Trees / Slide Double Left Rotation (DLR, left-right) Performed when C is unbalanced to the left (the left subtree is higher than the right subtree), A is right-heavy (the right subtree of A is 1 higher than the left subtree of A). Consists of a single left rotation at node A, followed by a single right rotation at node C. A SLR at A B T1 B A T T T T1 T C SLR at A C A T B T T1 B A T T T T1 T SRR at C B A C T1 TT T DLR = SLR + SRR AVL Trees / Slide Double Right Rotation (DRR, right-left) Performed when A is unbalanced to the right (the right subtree is higher than the left subtree), C is left-heavy (the left subtree of C is 1 higher than the right subtree of C). Consists of a single right rotation at node C, followed by a single left rotation at node A. A SRR at C T1 C B T A T1 B T C T T T T SLR at A B A C T1 TT T DRR = SRR + SLR AVL Trees / Slide 8 Insertion Analysis Insert the new key as a new leaf just as in ordinary binary search tree: O(logN) Then trace the path from the new leaf towards the root, for each node x encountered: O(logN) Check height difference: O(1) If satisfies AVL property, proceed to next node: O(1) If not, perform a rotation: O(1) The insertion stops when A single rotation is performed. Or, we ve checked all nodes in the path. Time complexity for insertion O(logN). logn AVL Trees / Slide 9 Deletion from AVL Trees Delete a node x as in ordinary binary search tree Note that the last node deleted is a leaf (case I in BST deletion) or a node with one child (case II in BST deletion). Then trace the path from the last node deleted towards the root (deletion path). For each node x encountered, check if heights of left(x) and right(x) differ by at most 1. If yes, proceed to parent(x). If no, perform an appropriate rotation at x. Continue to trace the path until we reach the root. Unlike insertion, more than one node may need rotation. AVL Trees / Slide Deletion Example Delete, Node 10 is unbalanced Single Rotation 8 0 0

6 AVL Trees / Slide 1 0 Cont d 0 0 AVL Trees / Slide Rotation in Deletion The rotation strategies (single or double) we learned can be reused here. Except for one new case: two subtrees of y are of the same height rotate with left child Continue to check parents 0 Oops!! Node 0 is unbalanced!! Single Rotation For deletion, after rotation, we need to continue tracing upward to see if AVL-tree property is violated at other nodes. rotate with right child AVL Trees / Slide AVL Trees / Slide Deletion Example Example (Cont d) Right most child of left subtree New case Double rotation