Announcements Assignment 2 is posted due Wed Oct

Announcements: Assignment #2 is posted, due Wed. Oct. 7 (no extension). Lab 4 is posted. No office hours Friday Oct. 2. Let me know if you have any suggestions for the Lab content. Take the code you handed in to Lab #4 (even if it did not work properly), do not change to the model solution and take my code. 1

Lecture 10: Binary Tree Sort http: //www. nevron. com/Gallery. Diagram. For. NET. Symmetrical. aspx 2

public class Split. List { Linked. List list 1; Linked. List list 2; Question #7: Assignment #1 public Linked. List join. List() { Linked. List restored; } list 1. rear. next= list 2. start; restored= new Linked. List(list 1. n + list 2. n, list 1. start, list 2. rear); return(restored); 3

public Split. List(int size 1, Linked. List list) { List. Node current, rear; int i; current= list. start; for (i=1; i < size 1; i++) { current= current. next; rear= current; current= current. next; rear. next= null; Question #6: Assignment #1 } list 1= new Linked. List(size 1, list. start, rear); list 2= new Linked. List(list. n-size 1, current, list. rear); } // A constructor for a Split. List 4

public void print. List(int n. Per. Line, int n. Digit) { List. Node current; int count=0; current= start; Question #5: while (current != null) Assignment #1 { count++; format_int(0, current. data, n. Digit); if (count % n. Per. Line == 0) System. out. println(); else System. out. print(" "); current= current. next; } if (count % n. Per. Line !=0) System. out. println(); } // This is in Linked. List class 5

public static void main(String args[]) Main: { int i; Linked. List list; Split. List split; Scanner in = new Scanner(System. in); Assignment list= new Linked. List(); while (list. read. Rear(in)) { System. out. println("The original list: "); list. print. List(5, 3); System. out. println("Starting the splits: "); for (i=1; i < list. n; i++) { split= new Split. List(i, list); System. out. println("Split List for " + i + " List 1: "); split. list 1. print. List(5, 3); System. out. println("Split List for " + i + " List 2: "); split. list 2. print. List(5, 3); list= split. join. List(); System. out. println("After restoration: "); list. print. List(5, 3); } } } #1 6

The original list: 0 9 1 8 2 7 3 6 4 5 Starting the splits: Split List for 1 List 1: 0 Split List for 1 List 2: 9 1 8 2 7 3 6 4 5 After restoration: 0 9 1 8 2 7 3 6 4 5 Split List for 2 List 1: 0 9 Split List for 2 List 2: 1 8 2 7 3 6 4 5 After restoration: 0 9 1 8 2 7 3 6 4 5 Split List for 3 List 1: 0 9 1 Split List for 3 List 2: 8 2 7 3 6 4 5 After restoration: 0 9 1 8 2 7 3 6 4 5 Begin. Max/Middle. Max/End. Max Split List for 4 List 1: 0 9 1 8 Split List for 4 List 2: 2 7 3 6 4 5 After restoration: 0 9 1 8 2 7 3 6 4 5 Split List for 5 List 1: 0 9 1 8 2 Split List for 5 List 2: 7 3 6 4 5 After restoration: 0 9 1 8 2 7 3 6 4 5 Split List for 6 List 1: 0 9 1 8 2 7 Split List for 6 List 2: 3 6 4 5 After restoration: 0 9 1 8 2 7 3 6 4 5 Split List for 7 List 1: 0 9 1 8 2 7 3 Split List for 7 List 2: 6 4 5 After restoration: 0 9 1 8 2 7 3 6 4 5 Split List for 8 List 1: 0 9 1 8 2 7 3 6 Split List for 8 List 2: 4 5 After restoration: 0 9 1 8 2 7 3 6 4 5 Split List for 9 List 1: 0 9 1 8 2 7 3 6 4 Split List for 9 List 2: 5 After restoration: 0 9 1 8 2 7 3 6 4 5 7

Binary search tree: If a node has key value x then -all keys in the left subtree have a data value which is less than or equal to x, and - all keys in the right subtree have a data value which is greater than x. 8

If items are inserted one by one, the resulting tree can have arbitrary shape: 9

http: //scienceblogs. com/goodmath/upload/2007/01/unbalanced-trees. jpg 10

Worst Case Shape Best Case Shape http: //www. brpreiss. com/books/opus 5/html/page 304. html 11

Inductive Definition of a Binary Search tree: A binary search tree is built up of nodes each of which has a key value, a left child and a right child (where the children are pointers to other nodes). [Basis] A tree with 0 nodes is a binary search tree with root NULL. [Inductive step] If T 1 and T 2 are binary search trees with roots r 1 and r 2 respectively, and r is a node with key value k, and further, all nodes in T 1 have key value ≤ k, and all nodes in T 2 have key value > k, then setting the left child of r to point to r 1 and the right child of r to point to r 2 creates a new binary search tree T with root r. T 1 and T 2 are said to be respectively the left and right subtrees of the tree T rooted at r. 12

The pseudocode for a binary tree sort is: Binary Tree Sort 1. Start with an empty binary search tree T. 2. Add each key to T. 3. Traverse T using an in order traversal. 13

In Order Traversals A traversal of the nodes of a binary search tree which visits the keys in sorted order is called an in order traversal of the tree. Recursive code for an in order traversal: /* root- pointer to the root node of a binary search tree. */ Inorder(root) 1. If (the left_child is not null) Inorder(root ->left_child) 2. Visit the root node. 3. If (the right_child is not null) Inorder(root->right_child) 14

public class Binary. Tree. Node { int data; Binary. Tree. Node left. Child; Binary. Tree. Node right. Child public void in. Order() { if (left. Child!= null) left. Child. in. Order(); System. out. print(data + " "); if (right. Child!= null) right. Child. in. Order(); } 15

public void pre. Order() { System. out. print(data + " "); if (left. Child!= null) left. Child. pre. Order(); if (right. Child!= null) right. Child. pre. Order(); } public void post. Order() { if (left. Child!= null) left. Child. post. Order(); if (right. Child!= null) right. Child. post. Order(); System. out. print(data + " "); } 16

Secret technique for doing traversals by hand: 17

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Constructor for a Binary. Tree. Node: public Binary. Tree. Node(int value) { data= value; left. Child= null; right. Child= null; } Next, code for adding a node to a binary tree which has at least one node in it already. 19

public void add. Tree(int value) { if (value <= data) { if (left. Child == null) left. Child= new Binary. Tree. Node(value); else left. Child. add. Tree(value); } else { if (right. Child == null) right. Child= new Binary. Tree. Node(value); else right. Child. add. Tree(value); } } 20

How much time does the binary tree sort take to sort: 1. in the best case? 2. in the worst case? 3. Give examples of input which can be provided for arbitrary n which realize the: 4. best case behaviour? 5. worst case behaviour? 6. How much extra space does it use: 7. in the best case? 8. in the worst case? 21