Recursion: Convert a given decimal number into binary using recursion

Problem Statement:

You are given a decimal number, you need to convert into binary number using recursion.

Example:

Input: 15
Output: 1111

Solution Explanation:

To get the binary number you need to use modulus operator “N%2”.

Then recursively call the function with the value (N/2).
The base case will be when “N==0”, we return from the function.

Time Complexity: O(logN)
Space Complexity: O(logN)

Code Solution

#include <iostream>
using namespace std;

void solution(int n)
{
    // base case
    if (n == 0) {
        cout << "0";
        return;
    }

    solution(n / 2);
    cout << n % 2;
}

int main()
{
    int n = 5;

    solution(n);

    return 0;
}

Output

0101