Print 1 to n using Recursion

Given an integer n. Print numbers from 1 to n using recursion.

Examples:

Input: n = 3
Output: [1, 2, 3]
Explanation: We have to print numbers from 1 to 3.

Input: n = 10
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Approach:

To solve this problem using recursion, we define a function that takes an integer n as an argument. The function first checks for a base case (n == 0) to stop the recursion. If not, it makes a recursive call with n - 1, ensuring smaller numbers are handled first. After the recursive call returns, the function prints the current value of n. This process continues until all numbers from 1 up to the given value are printed in order.

#include <iostream>
using namespace std;

void printNos(int n){
    
    // base condition
    if (n == 0)
        return;
        
    // recursive call
    printNos(n - 1);
    cout << n << " ";
}

int main(){
    int n = 3;
    printNos(n);
    return 0;
}

#include <stdio.h>

void printNos(int n){
    
    // base condition
    if (n == 0)
        return;
        
    // recursive call
    printNos(n - 1);
    printf("%d ", n);
}

int main(){
    int n = 3;
    printNos(n);
    return 0;
}

class GfG {
    static void printNos(int n){
        
        // base condition
        if (n == 0)
            return;
    
        // recursive call
        printNos(n - 1);
        System.out.print(n + " ");
    }
    public static void main(String[] args){
        int n = 3;
        printNos(n);
    }
}

def printNos(n):
    if n == 0:
        
        # base condition
        return
    
    # recursive call first
    printNos(n - 1)
  
    # print after recursion
    print(n, end=' ')


if __name__ == "__main__":
    n = 3
    printNos(n)

using System;

class GfG {
    static void printNos(int n){
        
        // base condition
        if (n == 0) {
            return;
        }
        
        // recursive call
        printNos(n - 1);
        Console.Write(n + " ");
    }

    static void Main(){
        int n = 3;
        printNos(n);
    }
}

function printNos(n){
    // base condition
    if (n == 0)
        return;
        
    // recursive call
    printNos(n - 1);
    
    // print in same line with space
    process.stdout.write(n + " ");
}

// Driver Code
let n = 3;
printNos(n);

1 2 3 

Time Complexity: O(n)
Auxiliary Space: O(n)