# Day 23: BST Level-Order Traversal | 30 Days Of Code | HackerRank Solution

Hello coders, today we are going to solve Day 23: BST Level-Order Traversal HackerRank Solution in C++, Java and Python.

Contents

## Objective

Today, we’re going further with Binary Search Trees.

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 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, 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.

## Constraints

• 1 <= N <= 20
• 1 <= node.data[i] <= 100

## 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.

Sample Input

``````6
3
5
4
7
2
1``````

Sample Output

``3 2 5 1 4 7 ``

Explanation

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.

## Solution – Day 23: BST Level-Order Traversal

### 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;
}```

### Java

```import java.util.*;
import java.io.*;
class Node{ Node left,right;
int data; Node(int data){ this.data=data; left=right=null;
}
}

class Solution{ static void levelOrder(Node root){
int height=getHeight(root);
for(int i=1;i<=height;i++) { printEachLevel(root,i);
}
} static void printEachLevel(Node root,int eachLevel)
{
if (root == null) return;
if (eachLevel == 1) System.out.print(root.data + " ");
else if (eachLevel > 1) { printEachLevel(root.left, eachLevel-1); printEachLevel(root.right, eachLevel-1); } }
static int getHeight(Node root){
if(root == null) { return 0; }
else
{
int lHeight= getHeight(root.left); int rHeight=getHeight(root.right); if(lHeight>rHeight) { return (lHeight+1); } else return (rHeight+1); } } public static 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; } } public static void main(String args[]){ Scanner sc=new Scanner(System.in); int T=sc.nextInt(); Node root=null; while(T-->0){ int data=sc.nextInt(); root=insert(root,data); } levelOrder(root); } }```

### Python

```import sys

class Node:
def __init__(self,data):
self.right=self.left=None
self.data = data
class Solution:
def insert(self,root,data):
if root==None:
return Node(data)
else:
if data<=root.data:
cur=self.insert(root.left,data)
root.left=cur
else:
cur=self.insert(root.right,data)
root.right=cur
return root
def levelOrder(self, root):
queue = [root]
while len(queue) is not 0:
curr = queue[0]
queue = queue[1:]
print(str(curr.data) + " ", end="")

if curr.left is not None:
queue.append(curr.left)
if curr.right is not None:
queue.append(curr.right)
T=int(input())
myTree=Solution()
root=None
for i in range(T):
data=int(input())
root=myTree.insert(root,data)
myTree.levelOrder(root)
```

Disclaimer: The above Problem (Day 23: BST Level-Order Traversal) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.