Sum of digit of a number using recursion

Given a number, we need to find sum of its digits using recursion.
Examples: 

Input : 12345
Output : 15

Input : 45632
Output :20
          

The step-by-step process for a better understanding of how the algorithm works. 
Let the number be 12345. 
Step 1-> 12345 % 10 which is equal-too 5 + ( send 12345/10 to next step ) 
Step 2-> 1234 % 10 which is equal-too 4 + ( send 1234/10 to next step ) 
Step 3-> 123 % 10 which is equal-too 3 + ( send 123/10 to next step ) 
Step 4-> 12 % 10 which is equal-too 2 + ( send 12/10 to next step ) 
Step 5-> 1 % 10 which is equal-too 1 + ( send 1/10 to next step ) 
Step 6-> 0 algorithm stops 
following diagram will illustrate the process of recursion 

C++

// Recursive C++ program to find sum of digits
// of a number
#include <bits/stdc++.h>
using namespace std;

// Function to check sum of digit using recursion
int sum_of_digit(int n)
{
    if (n == 0)
    return 0;
    return (n % 10 + sum_of_digit(n / 10));
}

// Driven code
int main()
{
    int num = 12345;
    int result = sum_of_digit(num);
    cout << "Sum of digits in "<< num
       <<" is "<<result << endl;
    return 0;
}

// THis code is contributed by
// SHUBHAMSINGH10

C

// Recursive C program to find sum of digits
// of a number
#include <stdio.h>

// Function to check sum of digit using recursion
int sum_of_digit(int n)
{
    if (n == 0)
       return 0;
    return (n % 10 + sum_of_digit(n / 10));
}

// Driven Program to check above
int main()
{
    int num = 12345;
    int result = sum_of_digit(num);
    printf("Sum of digits in %d is %d\n", num, result);
    return 0;
}

Java

// Recursive java program to
// find sum of digits of a number
import java.io.*;

class sum_of_digits
{
    // Function to check sum
    // of digit using recursion
    static int sum_of_digit(int n)
    {
        if (n == 0)
            return 0;
        return (n % 10 + sum_of_digit(n / 10));
    }

    // Driven Program to check above
    public static void main(String args[])
    {
        int num = 12345;
        int result = sum_of_digit(num);
        System.out.println("Sum of digits in " +
                           num + " is " + result);
    }
}

// This code is contributed by Anshika Goyal.

Python3

# Recursive Python3 program to
# find sum of digits of a number

# Function to check sum of
# digit using recursion
def sum_of_digit( n ):
    if n == 0:
        return 0
    return (n % 10 + sum_of_digit(int(n / 10)))

# Driven code to check above
num = 12345
result = sum_of_digit(num)
print("Sum of digits in",num,"is", result)

# This code is contributed by "Sharad_Bhardwaj".

C#

// Recursive C# program to
// find sum of digits of a number
using System;

class GFG {

    // Function to check sum
    // of digit using recursion
    static int sum_of_digit(int n)
    {
        if (n == 0)
            return 0;

        return (n % 10 + sum_of_digit(n / 10));
    }

    // Driven Program to check above
    public static void Main()
    {
        int num = 12345;
        int result = sum_of_digit(num);
        Console.WriteLine("Sum of digits in " +
                           num + " is " + result);
    }
}

// This code is contributed by Anant Agarwal.

PHP

<?php
// Recursive PHP program
// to find sum of digits
// of a number

// Function to check sum of
// digit using recursion
function sum_of_digit($n)
{
    if ($n == 0)
        return 0;
    return ($n % 10 +
            sum_of_digit($n / 10));
}

// Driven Code
$num = 12345;
$result = sum_of_digit($num);
echo("Sum of digits in " . $num . " is " . $result);

// This code is contributed by Ajit.
?>

Javascript

<script>

// Recursive Javascript program to find sum of digits
// of a number

// Function to check sum of digit using recursion
function sum_of_digit(n)
{
    if (n == 0)
    return 0;
    return (n % 10 + sum_of_digit(parseInt(n / 10)));
}

// Driven code
var num = 12345;
var result = sum_of_digit(num);
document.write( "Sum of digits in "+ num
   +" is "+result );

</script>

Output: 

Sum of digits in 12345 is 15

Besides writing (n==0 , then return 0) in the code given above we can also write it in this manner , there will be no change in the output .

if(n<10) return n; By writing this there will be no need to call the function for the numbers which are less than 10 

Time complexity : O(logn) where n is the given number. 

Auxiliary space : O(logn) due to recursive stack space.