Check if a number is binary or not in Java

Given a number N, the task is to check first whether the given number is binary or not and its value should be greater than 1. print true if N is the binary representation else print false.

Examples: 

Input: N = 1010 
Output: true 
Explanation: 
Given number is greater than 1 and none of its digits is greater than 1. Thus, it is a binary number greater than 1.

Examples: N = 1234 
Output: false 
Explanation: 
Given number is greater than 1 but some of its digits { 2, 3, 4} are greater than 1. Thus, it is not a binary number. 

Iterative Approach: 

1. Check if the number is less than or equal to 1. If it is then, print false.
2. Else if number is greater than 1 then, check if every digits of the number is 1 or 0.
3. If any digits of the number is greater than 1 then print false, else print true.

Below is the implementation of the above approach:

Java 

// Java program for the above approach
class GFG {

    // Function to check if number
    // is binary or not
    public static boolean isBinaryNumber(int num)
    {

        // Return false if a number
        // is either 0 or 1 or a
        // negative number
        if (num == 0 || num == 1
            || num < 0) {
            return false;
        }

        // Get the rightmost digit of
        // the number with the help
        // of remainder '%' operator
        // by dividing it with 10
        while (num != 0) {

            // If the digit is greater
            // than 1 return false
            if (num % 10 > 1) {
                return false;
            }
            num = num / 10;
        }
        return true;
    }

    public static void main(String args[])
    {
        // Given Number N
        int N = 1010;

        // Function Call
        System.out.println(isBinaryNumber(N));
    }
}