AVL Trees CSE 373 Data Structures

Binary Search Tree - Best Time All BST operations are O(d), where d is tree depth minimum d is for a binary tree with N nodes What is the best case tree? What is the worst case tree? So, best case running time of BST operations is O(log N) 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Binary Search Tree - Worst Time Worst case running time is O(N) What happens when you Insert elements in ascending order? Insert: 2, 4, 6, 8, 10, 12 into an empty BST Problem: Lack of “balance”: compare depths of left and right subtree Unbalanced degenerate tree 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Balanced and unbalanced BST 1 4 2 2 5 3 1 3 4 4 Is this “balanced”? 5 2 6 6 1 3 5 7 7 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Approaches to balancing trees Don't balance May end up with some nodes very deep Strict balance The tree must always be balanced perfectly Pretty good balance Only allow a little out of balance Adjust on access Self-adjusting 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Balancing Binary Search Trees Many algorithms exist for keeping binary search trees balanced Adelson-Velskii and Landis (AVL) trees (height-balanced trees) Splay trees and other self-adjusting trees B-trees and other multiway search trees 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Perfect Balance Want a complete tree after every operation tree is full except possibly in the lower right This is expensive For example, insert 2 in the tree on the left and then rebuild as a complete tree 6 5 Insert 2 & complete tree 4 9 2 8 1 5 8 1 4 6 9 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL - Good but not Perfect Balance AVL trees are height-balanced binary search trees Balance factor of a node height(left subtree) - height(right subtree) An AVL tree has balance factor calculated at every node For every node, heights of left and right subtree can differ by no more than 1 Store current heights in each node 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Height of an AVL Tree N(h) = minimum number of nodes in an AVL tree of height h. Basis N(0) = 1, N(1) = 2 Induction N(h) = N(h-1) + N(h-2) + 1 Solution (Fibonacci) N(h) > h (  1.62) h h-2 h-1 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Height of an AVL Tree N(h) > h (  1.62) Suppose we have n nodes in an AVL tree of height h. n > N(h) (because N(h) was the minimum) n > h hence log n > h (relatively well balanced tree!!) h < 1.44 log2n (i.e., O(height) takes O(logn)) 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Node Heights 6 6 4 9 4 9 1 5 1 5 8 Tree A (AVL) Tree B (AVL) height=3 balance factor=2-1=1 3 6 6 2 1 2 2 4 9 4 9 1 1 1 1 1 1 5 1 5 8 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Worst-case AVL Trees 4 nodes: h = 3 vs 3 7 nodes: h = 4 vs 3 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Node Heights after Insert 7 Tree A (AVL) Tree B (not AVL) balance factor 2-0 = 2 3 4 6 6 2 2 2 3 4 9 4 9 1 1 1 1 1 2 1 5 7 1 5 8 1 7 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Insert and Rotation in AVL Trees Insert operation may cause balance factor to become 2 or –2 for some node only nodes on the path from insertion point to root node have possibly changed in height So after the Insert, go back up to the root node by node, updating heights If a new balance factor (the difference hleft-hright) is 2 or –2, adjust tree by rotation around the node 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Single Rotation in an AVL Tree 3 3 6 6 2 3 2 2 4 9 4 8 1 1 1 2 1 1 1 1 5 8 1 5 7 9 1 7 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Insertions in AVL Trees Let the node that needs rebalancing be . There are 4 cases: Outside Cases (require single rotation) : 1. Insertion into left subtree of left child of . 2. Insertion into right subtree of right child of . Inside Cases (require double rotation) : 3. Insertion into right subtree of left child of . 4. Insertion into left subtree of right child of . The rebalancing is performed through four separate rotation algorithms. 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Insertion: Outside Case f Consider a valid AVL subtree b h G h h A D 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Insertion: Outside Case f Inserting into A destroys the AVL property at node f b h G h+1 h D A 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Insertion: Outside Case f Do a “right rotation” b h G h+1 h D A 1/15/2019 CSE 373 - AU 05 -- AVL Trees

f b G D A Single right rotation Do a “right rotation” h h+1 h 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Outside Case Completed b “Right rotation” done! (“Left rotation” is mirror symmetric) f h+1 h h A G D AVL property has been restored! 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Insertion: Inside Case f Consider a valid AVL subtree b h G h h A D 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Insertion: Inside Case f Inserting into D destroys the AVL property at node f Does “right rotation” restore balance? b h G h h+1 A D 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Insertion: Inside Case b “Right rotation” does not restore balance… now b is out of balance f h A h h+1 G D 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Insertion: Inside Case f Consider the structure of subtree Y… b h G h h+1 A D 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Insertion: Inside Case f Y = node i and subtrees V and W b h G d h h+1 A h or h-1 C E 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Insertion: Inside Case f We will do a left-right “double rotation” . . . b G d A C E 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Double rotation : first rotation left rotation complete d G b E C A 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Double rotation : second rotation f Now do a right rotation d G b E C A 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Double rotation : second rotation right rotation complete Balance has been restored d b f h h h or h-1 C A E G 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Example of Insertions in an AVL Tree 3 20 Insert 5, 40 1 2 10 30 1 1 25 35 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Example of Insertions in an AVL Tree 3 4 20 20 2 2 2 3 10 30 10 30 1 1 1 2 1 1 5 25 35 5 25 35 1 40 Now Insert 45 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Single rotation (outside case) 5 4 20 20 2 4 2 3 10 30 10 30 1 3 1 1 1 2 5 25 35 5 25 40 1 1 35 45 40 2 Imbalance 1 45 Now Insert 34 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Double rotation (inside case) 5 4 20 20 4 4 2 3 10 30 10 35 1 3 3 1 2 2 5 Imbalance 25 40 5 30 40 1 25 34 2 1 45 35 45 1 Insertion of 34 1 34 1/15/2019 CSE 373 - AU 05 -- AVL Trees

AVL Tree Deletion Similar but more complex than insertion Rotations and double rotations needed to rebalance Imbalance may propagate upward so that many rotations may be needed. 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Pros and Cons of AVL Trees Arguments for AVL trees: Search is O(log N) since AVL trees are always balanced. Insertion and deletions are also O(logn) The height balancing adds no more than a constant factor to the speed of insertion. Arguments against using AVL trees: Difficult to program & debug; more space for balance factor. Asymptotically faster but rebalancing costs time. Most large searches are done in database systems on disk and use other structures (e.g. B-trees). May be OK to have O(N) for a single operation if total run time for many consecutive operations is fast (e.g. Splay trees). 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 6 100 4 5 44 144 2 3 4 3 17 62 117 162 1 1 2 2 3 2 2 32 50 78 111 132 150 178 1 1 2 1 1 1 1 72 101 128 132 147 172 181 1 129 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 6 100 4 5 44 144 1 3 4 3 17 62 117 162 1 1 2 2 3 2 2 32 50 78 111 132 150 178 1 1 2 1 1 1 1 72 101 128 132 147 172 181 1 129 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 6 100 5 44 144 4 3 17 62 117 162 2 3 2 2 50 78 111 132 150 178 1 2 1 1 1 1 72 101 128 132 147 172 181 1 129 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 6 100 5 62 144 4 3 44 117 162 78 2 3 2 2 17 50 72 111 132 150 178 1 2 1 1 1 1 101 128 132 147 172 181 1 129 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 6 100 3 5 62 144 2 4 3 2 44 117 162 78 1 1 1 2 3 2 2 17 50 72 111 132 150 178 1 2 1 1 1 1 101 128 132 147 172 181 1 129 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 6 100 3 5 62 144 2 4 3 2 44 117 162 78 1 1 1 2 3 2 2 17 50 72 111 132 150 178 1 2 1 1 1 1 101 128 132 147 172 181 1 129 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 100 62 144 44 117 162 78 17 50 72 111 132 150 178 101 128 132 147 172 181 129 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 100 62 117 44 144 78 111 17 50 72 101 132 162 128 132 150 178 129 147 172 181 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 117 100 144 62 111 132 162 44 78 101 128 132 150 178 17 50 72 129 147 172 181 1/15/2019 CSE 373 - AU 05 -- AVL Trees

Deletion 5 117 4 4 100 144 3 2 3 3 62 111 132 162 2 2 1 2 1 2 2 44 78 101 128 132 150 178 1 1 1 1 1 1 1 17 50 72 129 147 172 181 1/15/2019 CSE 373 - AU 05 -- AVL Trees