problem no 63: level order traversal.

 Task

A level-order traversal, also known as a breadth-first search, visits each level of a tree's nodes from left to right, top to bottom. You are given a pointer, , pointing to the root of a binary search tree. Complete the levelOrder function provided in your editor so that it prints the level-order traversal of the binary search tree.

Hint: You'll find a queue helpful in completing this challenge.

Function Description

Complete the levelOrder function in the editor below.

levelOrder has the following parameter:
- Node pointer root: a reference to the root of the tree

Prints
- Print node.data items as space-separated line of integers. No return value is expected.

Input Format

The locked stub code in your editor reads the following inputs and assembles them into a BST:
The first line contains an integer,  (the number of test cases).
The  subsequent lines each contain an integer, , denoting the value of an element that must be added to the BST.

Constraints

Output Format

Print the  value of each node in the tree's level-order traversal as a single line of  space-separated integers.

Sample Input

6
3
5
4
7
2
1

Sample Output

3 2 5 1 4 7 

Explanation

The input forms the following binary search tree:

We traverse each level of the tree from the root downward, and we process the nodes at each level from left to right. The resulting level-order traversal is , and we print these data values as a single line of space-separated integers.

code:

void levelOrder(Node * root){

      queue<Node*> q;

      q.push(root);

      while(q.empty()==false)

      {

          Node * temp=q.front();

          cout<<temp->data<<" ";

          q.pop();

          if(temp->left!=NULL)

          {

              q.push(temp->left);

          }

          if(temp->right!=NULL)

          {

              q.push(temp->right);

          }

          

      }

      

    

  

    }