Day 23: BST Level-Order Traversal

Objective
Today, we're going further with Binary Search Trees. Check out the Tutorial tab for learning materials and an instructional video!

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.

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.

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.

Solution:

static void levelOrder(Node root)

    {

        //Write your code here

        Queue<Node> queue = new LinkedList<Node>();

        queue.add(root);

        while(queue.peek()!=null)

        {

        Node node =queue.remove();

        System.out.print(""+node.data+" ");

        if(node.left!=null)

        queue.add(node.left);

        if(node.right!=null)

        queue.add(node.right);

        }

    }