We recursively sum the digits of the number. If the result is a single digit, return it. Otherwise, recurse again on the sum.

1. Base case: If num < 10, return num.
2. Compute digit sum by taking num % 10 (last digit) + recursive call on num // 10.
3. If the digit sum is >= 10, recurse on the digit sum.

• Time Complexity: O(log N) per pass, O(log N) passes in worst case.
• Space Complexity: O(log N) (recursion stack).

def digital_root(num):
    if num < 10:
        return num
    
    def digit_sum(n):
        if n == 0:
            return 0
        return n % 10 + digit_sum(n // 10)
    
    return digital_root(digit_sum(num))