CSE 3358 NOTE SET 10 Data Structures and Algorithms

Complexity of Searching a Binary Tree  Worst Case: Linked List  O(n)  Best Case: Complete Binary Tree  O(lg n)  Average Case: Average Position in Average Tree  O(lg n)

Tree Traversal  Process of visiting every node exactly one time  No implied notion of order, so there are n! different traversals  Most are useless because  Depth-First  Breadth-First

Depth First Searching  In – order  Left  Visit  Right  Pre – Order  Visit  Left  Right  Post – Order  Left  Right  Visit

In - Order template void BST ::inorder(BSTNode *p) { if (p != 0) { inorder (p->left); visit(p); inorder (p->right); } In order Traversal:

Pre - Order template void BST ::preorder(BSTNode *p) { if (p != 0) { visit(p); preorder (p->left); preorder (p->right); } Pre order Traversal:

Post - Order template void BST ::postorder(BSTNode *p) { if (p != 0) { postorder (p->left); postorder (p->right); visit(p); } Post order Traversal:

Breadth-First Traversal  Visiting each node top-down-left-to-right  Use a queue to store the children of a node when the node is visited Breadth-First Traversal:

Breadth-First Traversal – Implementation Ideas

Breadth-First IMPL template void BST ::breadthFirst(BSTNode *p) { Queue *> queue; BSTNode *p = root; if (p != 0) { queue.enqueue(p); while(!queue.empty()) { p = queue.dequeue(); visit(p); if (p -> left != 0) queue.enqueue(p->left); if (p -> right != 0) queue.enqueue(p->right); }

To Recur or Not to Recur template void BST ::iterativePreorder() { Stack *> travStack; BSTNode *p = root; if (p != 0) { travStack.push(p); while(!travStack.empty()) { p = travStack.pop(); visit(p); if (p->right != 0) travStack.push(p->right); if (p->left != 0) travStack.push(p->left); }

Deleting  Three Cases: 1) Node is a leaf Set the pointer from the parent to null. release memory as appropriate. 2) Node has 1 child Set the pointer from the parent to the child of the node to be deleted release memory as appropriate 3) Node has 2 children Will require multi-step process

Deleting: Case Delete 3

Deleting: Case Delete 31

Deleting: Case 3 – Option 1: Deleting by Merging Delete Find the right-most node of the left subtree Make that node the parent of the right subtree (of the node to be deleted)

Deleting: Case 3 – Option 2: Deleting by Copying Delete Find the right-most node of the left subtree Copy the key of that node to the node to be deleted. Delete the node that was copied from.

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