Algorithms ROBERT SEDGEWICK | KEVIN WAYNE 3.2 BINARY SEARCH TREES ‣ BSTs ‣ iteration Algorithms ‣ ordered operations FOURTH EDITION ‣ deletion (see book or videos) ROBERT SEDGEWICK | KEVIN WAYNE https://algs4.cs.princeton.edu Last updated on 3/1/19 3:45 PM 3.2 BINARY SEARCH TREES ‣ BSTs ‣ iteration Algorithms ‣ ordered operations ‣ deletion (see book or videos) ROBERT SEDGEWICK | KEVIN WAYNE https://algs4.cs.princeton.edu Binary search trees Definition. A BST is a binary tree in symmetric order. root a left link a subtree A binary tree is either: Empty. ・ right child ・Two disjoint binary trees (left and right). of root null links Anatomy of a binary tree parent of A and R Symmetric order. Each node has a key, key left link S and every node’s key is: of E E X Larger than all keys in its left subtree. A R 9 value ・ C H associated ・Smaller than all keys in its right subtree. with R keys smaller than E keys larger than E Anatomy of a binary search tree 3 Differences between heaps and binary search trees Heap BST Supported insert, search, delete, Insert, delete-max operations ordered operations What is inserted Keys Key-value pairs Underlying Resizing array Linked nodes data structure Fixed shape given n Varies; Tree shape (complete binary tree) depends on data Somewhat ordered Totally ordered Ordering of keys parent > child for max heap left child < parent < right child Duplicate keys Yes No allowed? 4 Binary search trees: quiz 1 Which of the following properties hold? A. If a binary tree is heap ordered, then it is symmetrically ordered. B. If a binary tree is symmetrically ordered, then it is heap ordered. C. Both A and B. D. Neither A nor B. max key X X min key H H max key E E S S C C S S A C A R C HR H A E A R E XR X symmetricsymmetric order order heapheap order orderheap order symmetric order 5 BST representation in Java A BST contains a reference to a root Node. A Node is composed 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; BST private Value val; private Node left, right; Node key val public Node(Key key, Value val) { left right this.key = key; this.val = val; } BST with smaller keys BST with larger keys } Binary search tree Key and Value are generic types; Key is Comparable 6 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 slide */ } public Value get(Key key) { /* see next slide */ } public Iterable<Key> keys() { /* see slides in next section */ } public void delete(Key key) { /* see textbook */ } } 7 BST search (get) If less, go left; if greater, go right; if equal, search hit; if null node, search miss. H: search hit S E X A R G: search miss S C H E X M A R C H M 8 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 = 1 + depth of node. 9 BST insert (put) Associate value with key. inserting L S E X A R C H Search for key, then two cases: M Key in tree ⇒ reset value. search for L ends P ・ at this null link Key not in tree ⇒ add new node. S ・ E X A R C H M create new node L P S E X A R C H M reset links L P on the way up Insertion into a BST 10 BST insert: Java implementation Put. Associate value with key. inserting L Tr i c k y ! S E X A R public void put(Key key, Value val) C H { root = put(root, key, val); } M P private Node put(Node x, Key key, Value val) search for L ends at this null link { S if (x == null) return new Node(key, val); E X int cmp = key.compareTo(x.key); A R if (cmp < 0) x.left = put(x.left, key, val); C H else if (cmp > 0) x.right = put(x.right, key, val); M else if (cmp == 0) x.val = val; create new node L P return x; S } E X A R C H M reset links L P on the way up Cost. Number of compares = 1 + depth of node. Insertion into a BST 11 best case H C S Tree shape A E R X Many BSTs correspond to same set of keys. ・ typical case S Number of compares for bestsearch/insert case = 1 + depth of node. ・ H E X C S A R A E R X C H best case typical case A worst case H S C C S E X E A E R X A R H C H R S X typical case S A worst case E X C BST possibilities A R E C H H R S X BottomA line. worstTree case shape depends on order of insertion. C E BST possibilities H 12 R S X BST possibilities BST insertion: random order visualization Ex. Insert keys in random order. 13 Binary search trees: quiz 2 Suppose that you insert n keys in random order into a BST. What is the expected height of the resulting BST? P H T A. ~ lg n D O S U B. ~ ln n height A E I Y C. ~ 2 lg n D. ~ 2 ln n C M E. ~ 4.31107 ln n L 14 Correspondence between BSTs and quicksort partitioning P H T D O S U A E I Y C M L Remark. Correspondence is 1–1 if array has no duplicate keys. 15 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 into a BST in random order, the expected height is ~ 4.31107 ln n. How Tall is a Tree? expected depth of How Tall is a Tree? function-call stack in quicksort Bruce Reed Bruce Reed CNRS, Paris, France CNRS, Paris, France [email protected] [email protected] ABSTRACT purpose of this note to prove that for /3 -- ½ + ~,3 we ABSTRACT Let H~ be the height of a randompurpose binary of search this note tree toon prove n thathave: for /3 -- ½ + ~,3 we Let H~ be the height of a randomnodes. binary We search show thattree onthere n existshave: constants a = 4.31107... THEOREM 1. E(H~) = ~logn -/31oglogn + O(1) nodes. We show that there existsand/3 constants = 1.95... a = 4.31107... such that E(H~) = c~logn -/31oglogn + and Var(Hn) = O(1) . and/3 = 1.95... such that E(H~)O(1), = c~logn We also -/31oglogn show that +Var(H~) =THEOREM O(1). 1. E(H~) = ~logn -/31oglogn + O(1) and O(1), We also show that Var(H~) = O(1). Var(Hn) = O(1) . Remark By the definition of a, /3 = 3~ Categories and Subject Descriptors 7"g~" The first defi- Remark By the definition of a, nition/3 = 7"g~"given3~ Theis more first suggestive defi- of why this value is correct, Categories and Subject Descriptors as we will see. But… Worst-case height is n – 1. E.2 [Data Structures]: Trees nition given is more suggestive of why this value is correct, E.2 [Data Structures]: Trees as we will see. For more information on random binary search trees, one 1. THE RESULTS For more information on randommay binary consult search [6],[7], trees, [1], one [2], [9], [4], and [8]. may consult [6],[7], [1], [2], [9], [4],Remark and [8]. 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