Recitation 9 Tree Rotations and AVL Trees BSTs

Recitation 9 Tree Rotations and AVL Trees

BSTs Review: Binary Search Tree (BST) 1 ideal case 4 2 1 x < 4 2 5 3 worst case: O(n) lookup O(n) insertion O(n) deletion 3 x > 4 4 5

BSTs Make BSTs balanced! Balanced BST 4 2 1 worst case: O(log n) lookup O(log n) insertion O(log n) deletion 5 3 If a BST becomes unbalanced, we can rebalance it in O(log n).

BSTs Review: definition of Height public static int get. Height(Tree. Node t) { if (t == null) return -1; return 1 + Math. max(get. Height(t. left), get. Height(t. right)); } length of the longest path from a node to a leaf

BSTs Definition of Balanced public static boolean is. Balanced(Tree. Node t) { return t == null || Math. abs(get. Height(t. left) get. Height(t. right)) <= 1 && is. Balanced(t. left) && is. Balanced(t. right); } A tree is balanced if each of its subtrees is balanced and their heights differ by at most 1

BSTs is. Balanced: Recursion needed! 0 -1 All subtrees need to be balanced! -2 -3 -4 1 2 3 4

Tree Rotations

Tree Rotations Notation k+2 x Inorder traversal: A x B y C k+1 k y A k B k C Recall that the BST inorder traversal gives sorted order. A subtree of height k

Tree Rotations: Used to balance a BST k+2 x The blue pointers are the only ones that change. k+2 y k+1 k y x A B C k k+1 k k A Inorder traversals are the same B

Tree Rotations example 2 3 2 5 5 0 1 2 2 6 0 1 unbalanced node 1 3 Rotate 3 4 6 0 2 0 0 0 4

Tree Rotations Rebalancing unbalanced node k+3 k+2 y x k+2 k y x A k k+1 B k k A C k+1 B C

Tree Rotations Problem: Rotating a Zig-Zag! 3 unbalanced node 3 5 5 2 4 0 6 Rotate 2 2 0 6 left child taller 1 1 2 right child taller 0 3 Zig-Zag: taller children on opposite sides 4 0 3 We get the opposite Zig-Zag!

Tree Rotations Double rotate 3 unbalanced node 3 5 5 2 4 0 6 4 1 2 3 6 1 1 st Rotation 0 0 2 3 0 2 still unbalanced node

Tree Rotations Double rotate 3 5 2 2 4 1 3 0 2 5 0 6 1 0 3 2 nd Rotation 6 0 2 0 4

Tree Rotations Rebalancing with double rotate k+3 x 1 st Rotation k+2 k z A y A k+1 y k C D k+1 k k k B x z B k C k D

Tree Rotations Rebalancing with double rotate k+3 k+2 y x 2 nd Rotation k+2 k k+1 y z x A k+1 k z B k C k k k D A B k C k D

Tree Rotations Summary of Rotations Double rotation necessary Only single rotation necessary Balanced! x y x z A y z D B z x C B A C D Symmetry holds for the other cases B C D

Tree Rotations Question: What is the resulting tree? 3 5 0 2 2 6 0 1 1 3 0 4

Tree Rotations Question: What is the resulting tree? 3 3 5 5 0 2 2 6 0 1 2 3 1 2 0 4 6 1 1 st Rotation 3 0 1 0 0 4

Tree Rotations Question: What is the resulting tree? 3 2 5 3 2 0 3 1 2 1 0 2 6 0 4 2 nd Rotation 1 1 5 0 1 0 4 0 6

AVL Trees

AVL Trees Named after its two Soviet inventors, Georgy Adelson. Velsky and E. M. Landis, who described AVL tres in a paper in 1962. First invention of self-balancing BSTs. Later: red-black trees, splay trees, and others

AVL Trees AVL Tree: self-balancing BST 4 2 1 5 3 AVL invariant: the height difference between its left and right children is at most 1 Lookup works the same as a normal BST lookup worst case: O(log n) lookup O(log n) insertion O(log n) deletion

AVL Trees Inserting an element insert(E elem) 4 2 1 5 3 https: //www. cs. usfca. edu/~galles/visualization/A VLtree. html Insert like a normal BST and if the AVL invariant is broken, do a single or double rotation to fix it Localizing the problem: 1. Imbalance will occur only on the path from the root to the newly inserted node 2. Rebalancing should occur at the deepest node 3. Must search for possible imbalance all the way up to root

AVL Trees Why use AVL Trees? If Hash. Sets have a lookup of expected O(1), why use BSTs with an expected lookup time of O(log n)? Depends on the problem: 1. Binary Search Trees are great at keeping elements in sorted order. 2. Key Ranges: How many words in the set start with k and end in z? 3. find. Predecessor(E elem) and find. Successor(E elem) ● O(log n) for AVL Tree, expected case O(n) for Hash. Set 4. Better worst case lookup and insertion times

Prelim Information 1. Tree Rotations will not be tested on Prelim 2 1. You don’t need to be able to write Tree Rotations code but can find it online if interested