def powRecu(self, x, n):
        if n == 0:
            return 1.0
        
        if n % 2 == 0:
            return self.powRecu(x * x, n / 2)
        else:
            return x * self.powRecu(x * x, n / 2)
dataset/ETHPy150Open kamyu104/LeetCode/Python/powx-n.py/Solution.powRecu/original