Recursive program to check if number is palindrome or not

Difficulty Level : Easy
Last Updated : 22 Mar, 2021

Given a number, the task is to write a recursive function which checks if the given number is palindrome or not.
Examples:

Input : 121
Output : yes

Input : 532
Output : no

The approach for writing the function is to call the function recursively till the number is completely traversed from the back. Use a temp variable to store the reverse of the number according to the formula which has been obtained in this post. Pass the temp variable in the parameter and once the base case of n==0 is achieved, return temp which stores the reverse of a number.
Below is the implementation of the above approach:

C++

// Recursive C++ program to check if the
// number is palindrome or not
#include <bits/stdc++.h>
using namespace std;
 
// recursive function that returns the reverse of digits
int rev(int n, int temp)
{
    // base case
    if (n == 0)
        return temp;
 
    // stores the reverse of a number
    temp = (temp * 10) + (n % 10);
 
    return rev(n / 10, temp);
}
 
// Driver Code
int main()
{
 
    int n = 121;
     
    int temp = rev(n, 0);
   
    if (temp == n)
        cout << "yes" << endl;
    else
        cout << "no" << endl;
    return 0;
}

Java

// Recursive Java program to
// check if the number is
// palindrome or not
import java.io.*;
 
class GFG
{
 
// recursive function that
// returns the reverse of digits
static int rev(int n, int temp)
{
    // base case
    if (n == 0)
        return temp;
 
    // stores the reverse
    // of a number
    temp = (temp * 10) + (n % 10);
 
    return rev(n / 10, temp);
}
 
// Driver Code
public static void main (String[] args)
{
    int n = 121;
    int temp = rev(n, 0);
     
    if (temp == n)
        System.out.println("yes");
    else
        System.out.println("no" );
}
}
 
// This code is contributed by anuj_67.

Python3

# Recursive Python3 program to check
# if the number is palindrome or not
 
# Recursive function that returns
# the reverse of digits
def rev(n, temp):
 
    # base case
    if (n == 0):
        return temp;
 
    # stores the reverse of a number
    temp = (temp * 10) + (n % 10);
 
    return rev(n / 10, temp);
 
# Driver Code
n = 121;
temp = rev(n, 0);
 
if (temp != n):
    print("yes");
else:
    print("no");
 
# This code is contributed
# by mits

C#

// Recursive C# program to
// check if the number is
// palindrome or not
using System;
 
class GFG
{
 
// recursive function
// that returns the
// reverse of digits
static int rev(int n,
               int temp)
{
    // base case
    if (n == 0)
        return temp;
 
    // stores the reverse
    // of a number
    temp = (temp * 10) +
               (n % 10);
 
    return rev(n / 10, temp);
}
 
// Driver Code
public static void Main ()
{
    int n = 121;
    int temp = rev(n, 0);
     
    if (temp == n)
        Console.WriteLine("yes");
    else
        Console.WriteLine("no" );
}
}
 
// This code is contributed
// by anuj_67.

PHP

<?php
// Recursive PHP program to check
// if the number is palindrome or not
 
// Recursive function that returns
// the reverse of digits
function rev($n, $temp)
{
    // base case
    if ($n == 0)
        return $temp;
 
    // stores the reverse of a number
    $temp = ($temp * 10) + ($n % 10);
 
    return rev($n / 10, $temp);
}
 
// Driver Code
$n = 121;
$temp = rev($n, 0);
 
if ($temp != $n)
    echo "yes";
else
    echo "no";
 
// This code is contributed
// by Sach_Code
?>

Javascript

<script>
 
// Recursive Javascript program to check if the
// number is palindrome or not
 
// recursive function that returns the reverse of digits
function rev(n, temp)
{
    // base case
    if (n == 0)
        return temp;
 
    // stores the reverse of a number
    temp = (temp * 10) + (n % 10);
 
    return rev(Math.floor(n / 10), temp);
}
 
// Driver Code
 
    let n = 121;
     
    let temp = rev(n, 0);
     
    if (temp == n)
        document.write("yes" + "<br>");
    else
        document.write("no" + "<br>");
 
// This code is contributed by Mayank Tyagi
 
</script>

Output:

yes

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