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<class T>void BST<T>::inorder(BSTNode<T> *p) { if (p != 0) { inorder (p->left); visit(p); inorder (p->right); }}

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In order Traversal:

Pre - Order

template<class T>void BST<T>::preorder(BSTNode<T> *p) { if (p != 0) { visit(p); preorder (p->left); preorder (p->right); }}

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Pre order Traversal:

Post - Order

template<class T>void BST<T>::postorder(BSTNode<T> *p) { if (p != 0) { postorder (p->left); postorder (p->right); visit(p); }}

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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

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Breadth-First Traversal:

Breadth-First Traversal – Implementation Ideas

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Breadth-First IMPL

template<class T>void BST<T>::breadthFirst(BSTNode<T> *p) { Queue<BSTNode<T>*> queue; BSTNode<T> *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 <class T>void BST<T>::iterativePreorder(){ Stack<BSTNode<T>*> travStack; BSTNode<T> *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 1

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Delete 3

Deleting: Case 2

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Delete 31

Deleting: Case 3 – Option 1: Deleting by Merging

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Delete 25

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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

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Delete 25

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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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