Representation of an AVL Node class AVLNode { < declarations for info stored in node, e.g. int info; > AVLnode left; AVLnode right; int height; int height(AVLNode T) { return T == null? -1 : T.height; } … }; COSC 2P03 Week 5

AVL insert AVLNode insert(AVLNode T, AVLNode newNode) { if(T == null) T = newNode; else if(newNode.info < T.info) // insert in left subtree T.left = insert(T.left, newNode); if(height(T.left) - height(T.right) == 2) if(newNode.info < T.left.info) //left subtree of T.left T = rotateWithLeftChild(T); else //right subtree of T.left T = doubleWithLeftChild(T); } else // insert in right subtree T.right = insert(T.right, newNode); if(height(T.right) - height(T.left) == 2) if(newNode.info > T.right.info) // right subtree of T.right T = rotateWithRightChild(T); else // left subtree of T.right T = doubleWithRightChild(T); T.height = max(height(T.left), height(T.right)) + 1; return T; COSC 2P03 Week 5

Single rotation – left-left Insertion in left subtree of k2.left caused an imbalance at k2: need to rebalance from k2 down. AVLNode rotateWithLeftChild(AVLNode k2) { AVLNode k1 = k2.left; k2.left = k1.right; k1.right = k2; k2.height = max(height(k2.left, height(k2.right)) + 1; k1.height = max(height(k1.left), k2.height) + 1; return k1; } Note: right-right case is symmetric to above (left↔right). COSC 2P03 Week 5

Double rotation – right-left Insertion in right subtree of left child caused imbalance at k3: need to rebalance from k3 down. AVLNode doubleWithLeftChild(AVLNode k3) { k3.left = rotateWithRightChild(k3.left); return rotateWithLeftChild(k3); } Note: left-right case is symmetric to above (left↔right). COSC 2P03 Week 5

B Trees (section 4.7 of textbook) Commonly used for organizing data on disk Disk access time (time to read or write a block) dominates cost Very important to minimize number of disk accesses Some notes on terminology: This textbook uses the name B tree, but elsewhere they are known as B+ trees In this textbook, the order of a B tree is the maximum number of children per index node. Elsewhere, order refers to the minimum number of children in index nodes other than the root. COSC 2P03 Week 5

B tree definitions A B-tree of order M is an M-ary tree such that: Data items are only in the leaves Non-leaf (index) nodes store up to M-1 keys: key i determines the smallest possible key in subtree i+1. The root is either a leaf or has between 2 and M children Every node other than the root is at least half-full: All non-leaf nodes (except root) have at least M/2 children All leaves are at the same depth and have between L/2 and L records. COSC 2P03 Week 5

B tree and Binary Search Tree comparison Binary search trees: nodes have 0, 1 or 2 children and 1 key B-trees: non-leaf nodes have up to M children: a node with d keys has d+1 children Binary search trees: data is stored in both leaf and non-leaf nodes, and for any given index node N: N’s left subtree contains only items with keys < N’s key N’s right subtree contains only items with keys > N’s key B-trees: data is stored only in leaf nodes. Non-leaf nodes contain up to M-1 keys (k1, …, kM-1) and: Subtree to left of k1 contains only items with keys <k1 Subtree between ki and ki+1 contains only items with keys <ki+1 and ≥ki Subtree to right of kM-1 contains only items with keys ≥kM-1 COSC 2P03 Week 5

B-tree Example M=5 and L=3 100 150 200 10 30 50 110 120 130 160 180 220 240 260 3 10 30 50 100 110 120 130 150 160 180 200 220 240 260 7 15 35 80 105 115 125 140 155 165 190 210 225 250 270 20 90 170 230 275 COSC 2P03 Week 5

B-tree example: insert 40 100 150 200 10 30 50 110 120 130 160 180 220 240 260 3 10 30 50 100 110 120 130 150 160 180 200 220 240 260 7 15 35 80 105 115 125 140 155 165 190 210 225 250 270 20 40 90 170 230 275 COSC 2P03 Week 5

B-tree example: insert 70 100 150 200 10 30 50 80 110 120 130 160 180 220 240 260 3 10 30 50 80 100 110 120 130 150 160 180 200 220 240 260 7 15 35 70 90 105 115 125 140 155 165 190 210 225 250 270 20 40 170 230 275 COSC 2P03 Week 5

B-tree example: insert 25 30 100 150 200 10 20 50 80 110 120 130 160 180 220 240 260 3 10 20 30 50 80 100 110 120 130 150 160 180 200 220 240 260 7 15 25 35 70 90 105 115 125 140 155 165 190 210 225 250 270 40 170 230 275 COSC 2P03 Week 5

B-tree example: insert 235 30 100 150 200 10 20 50 80 110 120 130 160 180 220 230 240 260 3 10 20 30 50 80 100 110 120 130 150 160 180 200 220 230 240 260 7 15 25 35 70 90 105 115 125 140 155 165 190 210 225 235 250 270 40 170 275 COSC 2P03 Week 5

B-tree example: insert 280 150 30 100 200 240 10 20 50 80 110 120 130 160 180 220 230 260 275 3 10 20 30 50 80 100 110 120 130 150 160 180 200 220 230 240 260 275 7 15 25 35 70 90 105 115 125 140 155 165 190 210 225 235 250 270 280 40 170 COSC 2P03 Week 5

Heaps A heap is a binary tree that satisfies all of the following properties: Structure property: It is a complete binary tree Heap-order property: Every node satisfies the heap condition: The key of every node n must be smaller than (or equal to) the keys of its children, i.e. n.info  n.left.info and n.info  n.right.info COSC 2P03 Week 5