# Day 23 BST Level-Order Traversal Hackerrank Solution in C++

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## Problem Statement: BST Level-Order Traversal Hackerrank

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 root pointing to the root of a binary search tree. Complete the level order 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, T (the number of test cases).
• The T subsequent lines each contain an integer, data, denoting the value of an element that must be added to the BST.

### Output Format

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

6
3
5
4
7
2
1

3 2 5 1 4 7

### Explanation

The input forms the following binary search tree:

image Credit:- HackerRank
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 3–>2–>5–>1–>4–>7, and we print these data values as a single line of space-separated integers.

## Day 23 BST Level-Order Traversal Solution in C++

``````#include <iostream>
#include <cstddef>
#include <queue>
#include <string>
#include <cstdlib>
using namespace std;
class Node{
public:
int data;
Node *left,*right;
Node(int d){
data=d;
left=right=NULL;
}
};
class Solution{
public:
Node* insert(Node* root, int data){
if(root==NULL){
return new Node(data);
}
else{
Node* cur;
if(data<=root->data){
cur=insert(root->left,data);
root->left=cur;
}
else{
cur=insert(root->right,data);
root->right=cur;
}
return root;
}
}

void levelOrder(Node * root){
queue<Node *> q;
Node* n = root;

while(n != NULL){
cout << n->data << ' ';

if( n->left  != NULL ) q.push(n->left);
if( n->right != NULL ) q.push(n->right);
if( !q.empty() ) {
n = q.front();
q.pop();
} else {
n = NULL;
}
}
}

};//End of Solution
int main(){
Solution myTree;
Node* root=NULL;
int T,data;
cin>>T;
while(T-->0){
cin>>data;
root= myTree.insert(root,data);
}
myTree.levelOrder(root);
return 0;
}``````

Output:-

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