CS2012 Programming Techniques II

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1 10/02/2014 CS2012 Programming Techniques II Vasileios Koutavas Lecture 12 1

2 10/02/2014 Lecture 12 2

3 10/02/2014 Lecture 12 3 Answer until Thursday. Results Friday

4 10/02/2014 Lecture 12 4 Last Week Review: basic tree data structure Recursion: the simplest way to implement operations on trees Started: Binary Search Trees (BSTs) Today Recursive implementation of BST operations Insert search

5 10/02/2014 Lecture 12 5 Recursion We saw: gcd(x, y) fib(n) factorial(n) Also: converting recursive impl.!" iterative impl. The simplest implementations of tree operations are recursive.

6 10/02/2014 Lecture 12 6 Binary Trees A Bin Tree is a data structure Each node has 0 2 child nodes If 1 node: 1 node has no parent (root) All other nodes have exactly 1 parent

7 10/02/2014 Lecture 12 7 Binary Tree root height = 4 5 leafs size =11

8 10/02/2014 Lecture 12 8 Binary Tree its parent the node its right child its left child its left subtree its right subtree

9 10/02/2014 Lecture 12 9 Leaf 90 = 90 null null

10 10/02/2014 Lecture Empty tree null

11 10/02/2014 Lecture Printing a Tree Recursive implementations are the most natural 1. String tostring(treenode root) 2. { 3. if (root == null) return ; 4. else return tostring(root.left) root.data tostring(root.right); 7. } 1. class TreeNode<T> 2. { 3. T data; 4. TreeNode<T> left; 5. TreeNode<T> right; 6. } In-order traversal of the tree tostring "

12 10/02/2014 Lecture Binary Search Trees

13 10/02/2014 Lecture What is a Binary Search Tree? A BST is a Binary tree the key in each node is unique each node contains a key which is greater than keys in left subtree less than keys in right subtree (implied) Lexicographic ordering of keys 1. class BSTNode<K, V> 2. { 3. K key; 4. V value; 5. BSTNode<T> left; 6. BSTNode<T> right; 7. } A Bin Tree is a data structure Each node has 0 2 child nodes If 1 node: 1 node has no parent (root) All other nodes have exactly 1 parent

14 10/02/2014 Lecture Which of the following are BSTs? Violates the 3 rd property (ordering)

15 10/02/2014 Lecture Is this a BST? Violates the 2 nd property (uniqueness of keys) A BST is a Binary tree the key in each node is unique each node contains a key which is larger than keys in left subtree smaller than keys in right subtree

16 10/02/2014 Lecture Is this a BST? null A Bin Tree is a data structure Each node has 0 2 child nodes If 1 node: 1 node has no parent (root) All other nodes have exactly 1 parent A BST is a Binary tree the key in each node is unique each node contains a key which is larger than keys in left subtree smaller than keys in right subtree

17 10/02/2014 Lecture Cases when writing tree operations The tree is Empty Has one root with a left and a right subtree

18 10/02/2014 Lecture Following slides from Algorithms by Sedgewick & Wayne

19 3.2 BINARY SEARCH TREES Algorithms F O U R T H E D I T I O N BSTs ordered operations deletion R O B E R T S E D G E W I C K K E V I N W A Y N E Algorithms, 4 th Edition Robert Sedgewick and Kevin Wayne Copyright February 29, :50:38 AM

20 BSTs ordered operations deletion 2

21 Binary search trees Definition. A BST is a binary tree in symmetric order. root a left link A binary tree is either: Empty. Two disjoint binary trees (left and right). a subtree right child of root null links Anatomy of a binary tree Symmetric order. Each node has a key, and every node s key is: Larger than all keys in its left subtree. Smaller than all keys in its right subtree. left link of E parent of A and R E A C H R 9 S X key value associated with R keys smaller than E keys larger than E Anatomy of a binary search tree 3

22 BST representation in Java Java definition. A BST is a reference to a root Node. A Node is comprised of four fields: A Key and a Value. A reference to the left and right subtree. smaller keys larger keys private class Node { private Key key; private Value val; private Node left, right; public Node(Key key, Value val) { this.key = key; this.val = val; } } BST Node key left val right BST with smaller keys BST with larger keys Binary search tree Key and Value are generic types; Key is Comparable 4

23 BST implementation (skeleton) public class BST<Key extends Comparable<Key>, Value> { private Node root; root of BST private class Node { /* see previous slide */ } public void put(key key, Value val) { /* see next slides */ } public Value get(key key) { /* see next slides */ } public void delete(key key) { /* see next slides */ } public Iterable<Key> iterator() { /* see next slides */ } } 5

24 BST search and insert demo 6

25 BST search: Java implementation Get. Return value corresponding to given key, or null if no such key. public Value get(key key) { Node x = root; while (x!= null) { int cmp = key.compareto(x.key); if (cmp < 0) x = x.left; else if (cmp > 0) x = x.right; else if (cmp == 0) return x.val; } return null; } Cost. Number of compares is equal to 1 + depth of node. 7

26 BST insert Put. Associate value with key. Search for key, then two cases: Key in tree reset value. Key not in tree add new node. inserting L A search for L ends at this null link A C C create new node E E H H L M P M R P S R X S X S A C E reset links on the way up H L M P R X Insertion into a BST 8

27 BST insert: Java implementation Put. Associate value with key. public void put(key key, Value val) { root = put(root, key, val); } concise, but tricky, recursive code; read carefully! private Node put(node x, Key key, Value val) { if (x == null) return new Node(key, val); int cmp = key.compareto(x.key); if (cmp < 0) x.left = put(x.left, key, val); else if (cmp > 0) x.right = put(x.right, key, val); else if (cmp == 0) x.val = val; return x; } Cost. Number of compares is equal to 1 + depth of node. 9

28 Tree shape Many BSTs correspond to same set of keys. Number of compares for search/insert is equal to 1 + depth of node. best case A C E H R S X typical case A C E H R S X A C E worst case H R S X Remark. Tree shape depends on order of insertion. 10

29 BST insertion: random order visualization Ex. Insert keys in random order. 11

30 Correspondence between BSTs and quicksort partitioning K E S C I Q T A E M R U L P U O Remark. Correspondence is 1-1 if array has no duplicate keys. 12

31 BSTs: mathematical analysis Proposition. If N distinct keys are inserted into a BST in random order, the expected number of compares for a search/insert is ~ 2 ln N. Pf. 1-1 correspondence with quicksort partitioning. Proposition. [Reed, 2003] If N distinct keys are inserted in random order, expected height of tree is ~ ln N. How Tall is a Tree? Bruce Reed CNRS, Paris, France reed@moka.ccr.jussieu.fr ABSTRACT Let H~ be the height of a random binary search tree on n nodes. We show that there exists constants a = and/3 = such that E(H~) = c~logn -/31oglogn + O(1), We also show that Var(H~) = O(1). But Worst-case height is N. (exponentially small chance when keys are inserted in random order) 13

32 ST implementations: summary implementation guarantee average case search insert search hit insert ordered ops? operations on keys sequential search (unordered list) N N N/2 N no equals() binary search (ordered array) lg N N lg N N/2 yes compareto() BST N N 1.39 lg N 1.39 lg N? compareto() 14

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