Display nodes at given level in Binary Tree

Question:

Given a binary tree root node and a level, display all the nodes present in that level.

Example:

Consider the image given below:

Form the image above,

Nodes at level 1 are : 16
Nodes at level 2 are : 10, 25
Nodes at level 3 are : 7, 15, 18, 30

Solution:

This problem can be solved by recursive traversal, by using below logic:

   if(level == 1) 
   {
      print(node->data)
      return;
    }
    
    display_at_given_level(node->left, level - 1);
    display_at_given_level(node->right, level - 1);

 

Assume if you have only one node, and the level is 1.

Then it will satisfy the if condition and print the node value.

 

Suppose,
Now you have a tree as below and level is 2:

First you go to root node, from there you will recursively call “display_at_given_level” for the left node while reducing the given level.

Now, the if condition is met, and you will print the left node.

Similarly, you will call “display_at_given_level” for the right node while reducing the given level.

Now, the if condition is met, and you will print the right node.

Solution in C++

#include <iostream>
using namespace std;

// structure to hold binary tree node
struct Node
{
    int data;
    Node *left, *right;

    Node(int data)
    {
        this->data = data;
        this->left = this->right = nullptr;
    }
};


void display_at_given_level(Node* node, int level)  
{  
    //base case
    if (node == NULL) 
    {
      return;
    }
    
    //base case
    if(level == 1) 
    {
      cout<<node->data << " ";
      return;
    }
    
    display_at_given_level(node->left, level - 1);
    display_at_given_level(node->right, level - 1);
} 

int main()  
{  
    Node* root = nullptr;
    /* Binary tree:
              16
            /   \
           /     \
          10      25
         /  \    /  \
        /    \  /    \
       7    15 18     30

    */

    root = new Node(16);
    root->left = new Node(10);
    root->right = new Node(25);
    root->left->left = new Node(7);
    root->left->right = new Node(15);
    root->right->left = new Node(18);
    root->right->right = new Node(30);

    int level = 2;

    cout<<"Solution = " <<endl;
    display_at_given_level(root, level); 

    return 0;  
}  

 

Output:

Solution =
10 25