CS 2133: Data Structures

1. CS 2133: Data Structures Binary Search Trees

2. Definition of a Tree A Binary tree is a set T of nodes such that either 1. T is empty or 2. T is partitioned into three disjoint subsets. A single node r, the root Two (possibly empty) sets that are binary trees, called left and right subtrees of r. T T T Tl Tr

3. Structure of a Tree root struct Node{ int value; Node * left; Node * right; } 5 class Tree{ Node * root; public: Tree();//constructor Insert(int); Print(); } 4 6 3 10 7 9

4. Binary Search Trees • Binary Search Trees (BSTs) are an important data structure for manipulating dynamically changing data sets. • Each node has the following fields: • value: an identifying field inducing a total ordering • left: pointer to a left child (may be NULL) • right: pointer to a right child (may be NULL) • p: (sometimes) pointer to a parent node (NULL for root)

5. F B H A D K Binary Search Trees • BST property: Value of nodes in left subtree <= roots value Value of nodes in right subtree > roots value • Example:

6. Inorder Tree Walk • What does the following code do? TreeWalk(x) TreeWalk(left[x]); print(x); TreeWalk(right[x]); • A: prints elements in sorted (increasing) order • This is called an inorder tree walk • Preorder tree walk: print root, then left, then right • Postorder tree walk: print left, then right, then root

7. F B H A D K Inorder Tree Walk • Example: • How long will a tree walk take? • Prove that inorder walk prints in monotonically increasing order

8. Operations on BSTs: Search • Given a key and a pointer to a node, returns an element with that key or NULL: Node * TreeSearch(Node * ptr,int k){ if (ptr == NULL)return NULL; if (ptr->value==k) return ptr; if (k < ptr->value) return TreeSearch(ptr->left, k); else return TreeSearch(ptr->right, k); }

9. F B H A D K BST Search: Example • Search for D and C:

10. Operations of BSTs: Insert • Adds an element x to the tree so that the binary search tree property continues to hold • The basic algorithm • Like the search procedure above • Insert x in place of NULL • Use a “trailing pointer” to keep track of where you came from (like inserting into singly linked list)

11. F B H A D K BST Insert: Example • Example: Insert C C

12. Recursive Insert void BST<DataType>::InsertAux(BinNodePointer & locptr, const DataType & item) {if(locptr==0) locptr=new BSTNode(item); else if (item < locptr->data) InsertAux(locptr->left, item); else InsertAux(locptr->right, item); } Note &

13. BST Search/Insert: Running Time • What is the running time of TreeSearch() or insertAux()? • A: O(h), where h = height of tree • What is the height of a binary search tree? • A: worst case: h = O(n) when tree is just a linear string of left or right children • We’ll keep all analysis in terms of h for now • Later we’ll see how to maintain h = O(lg n)

14. BST Destroy the tree BinaryTree::~BinaryTree(){destroyTree(root)} Void BinaryTree::destroyTree(TreeNode *& treePrt) { if (tree{tr != NULL;) { destroyTree(treePtr->leftChildPtr); destroyTree(treePtr->rightChildPtr); delete treePtr; treePtr=NULL; } }

15. F Example: delete Kor H or B B H C A D K BST Operations: Delete • Deletion is a bit tricky • 3 cases: • x has no children: • Remove x • x has one child: • Splice out x • x has two children: • Swap x with successor • Perform case 1 or 2 to delete it

16. Delete Algorithm void TreeDelete(Node *& root, int x) { if(root!=NULL){ if(x<root->val)TreeDelete(root->left, x); else if(x>root->val)TreeDelete(root->right, x); else { // we have found it so lets delete it // tree points at it right!? // If No children we just delete it if(root->left==NULL && root->right==NULL){ delete(root); root=NULL; return; }

17. Delete Continued with one child // Check to see if the node to delete has only one child if(root->left==NULL){ // Then splice in the right side tree_ptr=root->right; delete root; root=tree_ptr; }else if (root->right==NULL){ // splice left tree_ptr=root->left; delete(root); root=tree_ptr; } else // both children exist! so...

18. Delete Continued with two children // Here root has two children // We first find the successor { tempPtr=root->right; // Go right // and the all the way to the left while(tempPtr->left!=NULL){ predPtr=tempPtr; tempPtr=tempPtr->left; } root->val=tempPtr->val; // Copy the value up to the node if(root->right==tempPtr)root->right=tempPtr->right; else predPtr->left=tempPtr->right; delete tempPtr; } root pred temp

19. BST Operations: Delete • Why will case 2 always go to case 0 or case 1? • A: because when x has 2 children, its successor is the minimum in its right subtree • Could we swap x with predecessor instead of successor? • A: yes. Would it be a good idea? • A:See the Eppinger paper!!

20. Sorting With Binary Search Trees • Informal code for sorting array A of length n: BSTSort(A) for i=1 to n TreeInsert(A[i]); InorderTreeWalk(root); • What will be the running time in the • Worst case? • Average case?

21. F B H C A D K Left Child Right Sibling Representation F B H A D H C RCLS format Normal format Really is a binary way to represent an n ary tree. Also each child has a pointer to its parent.

22. Recursive Traversal of a LCRS Tree F Traverse(t) { if (t != NULL) then Traverse(t.child()) If (t.sibling() !=NULL) then Traverse(t.sibling()) } B H A D H C

23. Algorithm BT->LCRS Format Can you design an algorithm that will convert a binary tree to the left child right sibling format?

24. Saving a binary search tree in a file • We can do this so the tree gets restored to the original shape. How do we do this? save in preorder ! • Or we can restore it in a balanced shape save in inorder and ?

25. Loading a balanced tree froma sorted list readTree(Node *& treePtr,int n) { if (n>0) { // read in the left subtree treePtr=new Node(NULL,NULL); readTree(treePtr->left,n/2); cin>> treePtr->val; // read in the right subtree readTree(treePtr->right,(n-1)/2); }

26. F B H A D K Level Order Traversal How do you perform an level by level traversal? Q.enq(root) While(Q not empty){ x = Q.deq(); Print x; Q.enq(left_child); Q.enq(right_child); } NOTE: This uses aqueue as its support data structure

27. Huffman Codes Variable length coding schemes. These schemes represent characters that occur more frequently with shorter codes. We thus attempt to minimize the expected length of the code for the character. Expected length = The weights represent the probability of occurrence of the character within the coded string. It is a value between 0 and 1.

28. Constructing the Code 1. Initialize a list of one-node binary trees containing the weights of each char 2. Do the following n-1 times Find two trees T’ and T’’ in this list with minimal weight Replace these two trees with a binary tree whose root is w’+w’’, and label the pointers to these trees with 0 and 1 W’+w’’ 1 0 T’’ T’

29. Example Construction 1.0 1 Also,this is Immediately Decodable! A=00 B=01 C=10 D=110 E=1110 F=1111 .40 1 0 .20 0 1 0 .60 .10 0 1 0 1 .35 .25 .20 .10 .08 .02 A B C D E F

30. Expression Trees + * + + X 3 Y W 5 W + 5 * X + 3 + Y inorder traversal ((W+5)*X + (3+Y)) with parens How about preorder and postorder traversals?

31. Problem A preorder traversal of a binary tree produced ADFGHKLPQRWZ and an inorder traversal produced GFHKDLAWRQPZ Draw the binary tree.

32. Program to Build Tree void BuildTree(string in,string post) { char root; int len=in.length(); // NOTE: Last char of postorder traveral is the root of the tree if(len==0)return; root=post[len-1]; cout<<root; // print the root // Search for root in the inorder traversal list int i=0; while(in[i]!=root)i++; // skip to the root // i now points to the root in the inorder traversal // The chars from 0 to i-1 are in the left subtree and // the chars from i+1 to len_in-1 are in the right sub tree. // Process left sub tree BuildTree(in.substr(0,i),post.substr(0,i)); //Process right sub tree BuildTree(in.substr(i+1,len-i-1), post.substr(i,len-i-1)); } 0 1 2 . . . substr(start, len)