Binary trees. Binary Tree. A binary tree is a tree where each node has at most two children The two children are ordered ( left, right ) 4/23/2013

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1 Binary trees Binary Tree A binary tree is a tree where each node has at most two children The two children are ordered ( left, right ) Right sub-tree vs. Left sub-tree 2 1

2 Balanced trees (Height-)balanced trees The left and right sub-trees heights differ by at most one The two sub-trees are (height-)balanced Perfectly balanced nodes Several alternative definitions 3 Complete trees Complete binary tree Every level, except possibly the last, is completely filled, and all nodes are as far left as possible 4 2

3 Tree representation A complete tree can be efficently stored in an array 5 Degenerate trees 6 3

4 Traversal in binary trees Pre-order process root node, then its left/right sub-trees In-order process left sub-tree, then root node, then right Post-order process left/right sub-trees, then root node 7 Traversal in binary trees Root pre-order: in-order: post-order:

5 Traversal trick To quickly generate a traversal: Trace a path around the tree As you pass a node on the proper side, process it pre-order: left side in-order: bottom post-order: right side Root pre-order: in-order: post-order: Exercise Give pre-, in-, and post-order traversals for the following tree: Root 42 pre: in: post:

6 Polish prefix notation Akas: Polish notation, prefix notation Created in 1924 by the Polish logician Jan Łukasiewicz Operators are on the left of their operands If the arity of the operators is fixed no need for parentheses or other brackets E.g.: 3 * ( ) * ( x + y ) / ( 2 - z ) / + x y - 2 z 11 Reverse Polish notation (Re-)Invented by Bauer and Dijkstra in early 1960s to exploit stack for evaluating expressions Operator follows all of its operands If the arity of the operators is fixed no need for parentheses or other brackets E.g.: 3 * ( ) * ( x + y ) / ( 2 - z ) x y + 2 z - / 12 6

7 Traversals and notations In-order: ( ) * 2 + ( 6-3 ) Pre-order: + * Post-order: * An oversimplified binary tree public class BTree<T> { public BNode<T> root; public BTree() { root = new BNode<T>(); public void InOrder() { root.inorder(); public void PreOrder() { root.preorder(); public void PostOrder() { root.postorder(); 7

8 An oversimplified binary tree public class BNode<T> { T data; BNode<T> parent; BNode<T> left; BNode<T> right; public BNode() { public BNode(T d) { this(); data = d; public BNode(T d, BNode<T> l, BNode<T> r) { this(d); addleftchild(l); addrightchild(r); [ ] 15 An oversimplified binary tree [ ] [ ] public void addleftchild(bnode<t> n) { n.parent = this; left = n; public void addrightchild(bnode<t> n) { n.parent = this; right = n; 16 8

9 An oversimplified binary tree [ ] [ ] public void removeleftchild(bnode<t> n) { left = null; public void removerightchild(bnode<t> n) { right = null; 17 Traversal [ ] [ ] public void InOrder() { System.out.print("("); if(left!= null) left.inorder(); System.out.print(data); if(right!= null) right.inorder(); System.out.print(")"); 18 9

10 Traversal [ ] [ ] public void PreOrder() { System.out.print(data); System.out.print(" "); if(left!= null) left.preorder(); if(right!= null) right.preorder(); 19 Traversal [ ] public void PostOrder() { if(left!= null) left.postorder(); if(right!= null) right.postorder(); System.out.print(data); System.out.print(" "); 20 10

11 Example 21 BST Binary Search Tree 11

12 Binary search trees A binary tree where each non-empty node R has the following properties: Elements of R s left sub-tree contain data less than R s data Elements of R s right sub-tree contain data greater than R s R s left and right sub-trees are also binary search trees root Binary search trees BSTs store their elements in sorted order, which is helpful for searching/sorting tasks 24 12

13 Excercise Is it a legal binary search tree? Excercise Is it a legal binary search tree?

14 Excercise Is it a legal binary search tree? Excercise Is it a legal binary search tree? m g q b k x e 28

15 Excercise Is it a legal binary search tree? Searching in a BST Describe an algorithm for searching a binary search tree (try searching for 31, then 6) root

16 Searching in a BST Searching in a BST is If the tree is balanced, then Searching for an element is 31 Showdown Array List Hash BST add(element) O(1) O(1) O(1) O(ln n) remove(object) O(n) + O(n) O(n) + O(1) O(1) O(ln n) get(index) O(1) O(n) n.a. n.a. set(index, element) O(1) O(n) + O(1) n.a. n.a. add(index, element) O(1) + O(n) O(n) + O(1) n.a. n.a. remove(index) O(n) O(n) + O(1) n.a. n.a. contains(object) O(n) O(n) O(1) O(ln n) indexof(object) O(n) O(n) n.a. n.a

17 Minimalistic BST public class BstNode { int key; String data; BstNode parent; BstNode left; BstNode right; 33 Operations on BST Searching for a key Inserting a node Find sub-tree minimum/maximum Find node successor/predecessor Delete a node 34 17

18 Searching for a key public BstNode search(int k) { if(key == k) return this; else if(key > k && left!= null) return left.search(k); else if(key < k && right!= null) return right.search(k); else return null; Inserting a node

19 Inserting a node protected void add(bstnode x) { if(root == null) { root = x; else { BstNode n = root, m = root; while(n!= null) { m = n; if(x.key < n.key) n = n.left; else n = n.right; if(x.key < m.key) m.addleftchild(x); else m.addrightchild(x); 37 Add Left/Right child public void addleftchild(bstnode n) { n.parent = this; this.left = n; public void addrightchild(bstnode n) { n.parent = this; this.right = n; 38 19

20 Sub-tree minimum Sub-tree minimum

21 Sub-tree maximum Sub-tree maximum

22 Sub-tree minimum/maximum public BstNode getmin() { BstNode m = this; while(m.left!= null) m = m.left; return m; public BstNode getmax() { BstNode m = this; while(m.right!= null) m = m.right; return m; 43 Successor

23 Successor If a node has a right child, the successor is the minimum element of the right sub-tree Successor

24 Successor If a node has no right child, the successor is the first ancestor whose left sub-tree the node is in Node successor public BstNode getsuccessor() { if(this.right!= null) { return this.right.getmin(); else { BstNode n = this; while(n.isrightchild()) n = n.parent; return n.parent; 48 24

25 Node successor public boolean isrightchild() { return parent!= null && parent.right == this; 49 Predecessor

26 Predecessor Delete node (leaf)

27 Delete node (leaf) protected void _deletenodeleaf() { if(isleftchild()) parent.left = null; else parent.right = null; 53 Delete node (node with single child)

28 Delete node (node with single child) protected void _deletenode1child() { BstNode c = left!= null? left : right; if(isleftchild()) parent.addleftchild(c); else parent.addrightchild(c); Delete node (node with two children)

29 Delete node (node with two children) protected void _deletenode2children() { BstNode t = getsuccessor(); key = t.key; data = t.data; t.deletenode(); 57 Delete node public void deletenode() { if(left == null && right == null) _deletenodeleaf(); else if(left == null right == null) _deletenode1child(); else _deletenode2children(); 58 29

30 Complexity & tree balancing Almost all operations in a BST are Rule of thumb: Everything run smoothly while Tree balancing Rotations Self-balancing trees Red-black tree Scapegoat tree AVL tree 59 Licenza d uso Queste diapositive sono distribuite con licenza Creative Commons Attribuzione - Non commerciale - Condividi allo stesso modo (CC BY-NC-SA) Sei libero: di riprodurre, distribuire, comunicare al pubblico, esporre in pubblico, rappresentare, eseguire e recitare quest'opera di modificare quest'opera Alle seguenti condizioni: Attribuzione Devi attribuire la paternità dell'opera agli autori originali e in modo tale da non suggerire che essi avallino te o il modo in cui tu usi l'opera. Non commerciale Non puoi usare quest'opera per fini commerciali. Condividi allo stesso modo Se alteri o trasformi quest'opera, o se la usi per crearne un'altra, puoi distribuire l'opera risultante solo con una licenza identica o equivalente a questa