We repeatedly divide the number by 2 and build the binary string from the remainders using recursion.

1. Base case: If n == 0, return "0". If n == 1, return "1".
2. Recursive step: Return decimal_to_binary(n // 2) + str(n % 2).

• Time Complexity: O(log N).
• Space Complexity: O(log N) (recursion stack).

def decimal_to_binary(n):
    if n == 0:
        return "0"
    if n == 1:
        return "1"
    return decimal_to_binary(n // 2) + str(n % 2)