3.2 BINARY SEARCH TREES. BSTs ordered operations iteration deletion. Algorithms ROBERT SEDGEWICK KEVIN WAYNE.

Size: px

Start display at page:

Download "3.2 BINARY SEARCH TREES. BSTs ordered operations iteration deletion. Algorithms ROBERT SEDGEWICK KEVIN WAYNE."

Philomena Stokes
6 years ago
Views:

1 3.2 BINY T lgorithms BTs ordered operations iteration deletion OBT DGWIK KVIN WYN

2 Binary search trees Definition. BT is a binary tree in symmetric order. binary tree is either: a subtree a left link root mpty. Two disjoint binary trees (left and right). ymmetric order. ach node has a key, and every node s key is: Larger than all keys in its left subtree. maller than all keys in its right subtree. left link of parent of and null links 9 right child of root key value associated with keys smaller than keys larger than 3

3 Binary search tree demo earch. If less, go left; if greater, go right; if equal, search hit. successful search for 4

4 Binary search tree demo Insert. If less, go left; if greater, go right; if null, insert. insert G G 5

5 BT representation in Java Java definition. BT is a reference to a root Node. Node is composed of four fields: Key and a Value. reference to the left and right subtree. smaller keys larger keys private class Node { private Key key; private Value val; private Node left, right; Node key BT val } public Node(Key key, Value val) { this.key = key; this.val = val; } left right BT with smaller keys BT with larger keys Binary search tree Key and Value are generic types; Key is omparable 6

6 BT implementation (skeleton) public class BT<Key extends omparable<key>, Value> { private Node root; root of BT 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> iterator() { /* see slides in next section */ } public void delete(key key) { /* see textbook */ } } 7

7 BT search: Java implementation Get. eturn 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; } ost. Number of compares = 1 + depth of node. 8

8 BT insert Put. ssociate value with key. earch for key, then two cases: Key in tree reset value. Key not in tree add new node. inserting L search for L ends at this null link create new node L P P reset links on the way up L P Insertion into a BT 9

9 BT insert: Java implementation Put. ssociate value with key. public void put(key key, Value val) { root = put(root, key, val); } 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; } Warning: concise but tricky code; read carefully! ost. Number of compares = 1 + depth of node. 10

10 any BTs correspond to same set of keys. Number of compares for search/insert = 1 + depth of node. Bottom line. Tree shape depends on order of insertion. 11 Tree shape best case typical case worst case

11 BT insertion: random order visualization x. Insert keys in random order. 12

12 Binary search trees: quiz 1 What is the expected number of compares to sort n distinct keys using the following sorting algorithm? 1. huffle the keys. 2. Insert the keys into a BT, one at a time. 3. Do an inorder traversal of the BT.. ~ n lg n B. ~ n ln n. ~ 2 n lg n D. ~ 2 n ln n 13

13 orrespondence between BTs and quicksort partitioning P T D O U I Y L emark. orrespondence is 1 1 if array has no duplicate keys. 14

14 BTs: mathematical analysis Proposition. If n distinct keys are inserted into a BT in random order, the expected number of compares for a search/insert is ~ 2 ln n. Pf. 1 1 correspondence with quicksort partitioning. Proposition. [eed, 2003] If n distinct keys are inserted into a BT in random order, the expected height is ~ ln n. expected depth of function-call stack in quicksort ow Tall is a Tree? Bruce eed N, Paris, France reed@moka.ccr.jussieu.fr BTT Let ~ be the height of a random binary search tree on n nodes. We show that there exists constants a = and/3 = such that (~) = c~logn -/31oglogn + O(1), We also show that Var(~) = O(1). But Worst-case height is n 1. [ exponentially small chance when keys are inserted in random order ] 15

15 T implementations: summary implementation guarantee average case search insert search hit insert operations on keys sequential search (unordered list) n n n n equals() binary search (ordered array) log n n log n n compareto() BT n n log n log n compareto() Why not shuffle to ensure a (probabilistic) guarantee of log n? 16

16 3.2 BINY T lgorithms BTs iteration ordered operations deletion OBT DGWIK KVIN WYN

17 Binary search trees: quiz 2 In which order does traverse(root) print the keys in the BT? private void traverse(node x) { if (x == null) return; traverse(x.left); tdout.println(x.key); traverse(x.right); } root. B.. D. 18

18 Inorder traversal inorder() inorder() inorder() print inorder() print done done print inorder() inorder() print inorder() print done done print done done print inorder() print done done output: 19

19 Inorder traversal Traverse left subtree. nqueue key. Traverse right subtree. public Iterable<Key> keys() { Queue<Key> q = new Queue<Key>(); inorder(root, q); return q; } key BT val private void inorder(node x, Queue<Key> q) { if (x == null) return; inorder(x.left, q); q.enqueue(x.key); inorder(x.right, q); } BT with smaller keys smaller keys, in order left right BT with larger keys key larger keys, in order all keys, in order Property. Inorder traversal of a BT yields keys in ascending order. 20

20 unning time Property. Inorder traversal of a BT takes linear time. ilicon Valley, eason 4, pisode 5 21

21 LVL-OD TVL Level-order traversal of a binary tree. Process root. Process children of root, from left to right. Process grandchildren of root, from left to right. T level-order traversal: T 22

22 LVL-OD TVL Q2. Given the level-order traversal of a BT, how to (uniquely) reconstruct? x. T T 24

23 3.2 BINY T lgorithms BTs iteration ordered operations deletion OBT DGWIK KVIN WYN

24 inimum and maximum inimum. mallest key in BT. aximum. Largest key in BT. min () m max Q. ow to find the min / max? 26

25 Floor and ceiling Floor. Largest key in BT query key. eiling. mallest key in BT query key. floor(g) () m ceiling(q) floor(d) Q. ow to find the floor / ceiling? 27

26 omputing the floor Floor. Largest key in BT k? Key idea. To compute floor(key), search for key. On search path, must encounter floor(key) and ceiling(key). Why? floor(g) () m floor(d) 28

27 omputing the floor public Key floor(key key) { return floor(root, key, null); } key in node is too big (so look in left subtree) private Key floor(node x, Key key, Key best) { if (x == null) return best; int cmp = key.compareto(x.key); if (cmp < 0) return floor(x.left, key, best); else if (cmp > 0) return floor(x.right, key, x.key); else if (cmp == 0) return x.key; } key in node is best candidate for floor (but maybe better one in right subtree) 29

28 ank and select ank. ow many keys < key? elect. Key of rank k. Q. ow to implement rank() and select() efficiently for BTs?. In each node, store the number of nodes in its subtree. subtree count

29 BT implementation: subtree counts private class Node { private Key key; private Value val; private Node left; private Node right; private int count; } public int size() { return size(root); } private int size(node x) { if (x == null) return 0; return x.count; ok to call } when x is null number of nodes in subtree private Node put(node x, Key key, Value val) initialize subtree { count to 1 if (x == null) return new Node(key, val, 1); 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; x.count = 1 + size(x.left) + size(x.right); } return x; 31

30 omputing the rank ank. ow many keys < key? ase 1. [ key < key in node ] Keys in left subtree? count Key in node? 0 Keys in right subtree? 0 node count ase 2. [ key > key in node ] Keys in left subtree? all Key in node. 1 Keys in right subtree? count ase 3. [ key = key in node ] Keys in left subtree? count Key in node. 0 Keys in right subtree? 0 32

31 ank ank. ow many keys < key? asy recursive algorithm (3 cases!) node count public int rank(key key) { return rank(key, root); } private int rank(key key, Node x) { } if (x == null) return 0; int cmp = key.compareto(x.key); if (cmp < 0) return rank(key, x.left); else if (cmp > 0) return 1 + size(x.left) + rank(key, x.right); else if (cmp == 0) return size(x.left); 33

32 BT: ordered symbol table operations summary sequential search binary search BT search n log n h insert n n h min / max n 1 h h = height of BT floor / ceiling n log n h rank n log n h select n 1 h ordered iteration n log n n n order of growth of running time of ordered symbol table operations 34

33 T implementations: summary implementation guarantee average case search insert search hit insert ordered ops? key interface sequential search (unordered list) n n n n equals() binary search (ordered array) log n n log n n compareto() BT n n log n log n compareto() red-black BT log n log n log n log n compareto() Next week. Guarantee logarithmic performance for all operations. 35

Algorithms. Algorithms 3.2 BINARY SEARCH TREES. BSTs ordered operations iteration deletion (see book or videos) ROBERT SEDGEWICK KEVIN WAYNE

Algorithms. Algorithms 3.2 BINARY SEARCH TREES. BSTs ordered operations iteration deletion (see book or videos) ROBERT SEDGEWICK KEVIN WAYNE lgorithms OBT DGWIK KVIN WYN 3.2 BINY T lgorithms F O U T D I T I O N BTs ordered operations iteration deletion (see book or videos) OBT DGWIK KVIN WYN https://algs4.cs.princeton.edu Last updated on 10/9/18