Pow(x, n) LeetCode solution in Python

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class Solution:
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    def myPow(self, x: float, n: int) -> float:
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        self.x = x
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        if n == 0:
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            return 1
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        isInverted = False
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        if n < 0:
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            isInverted = True
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            n = -1 * n
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        result = self.pow(n)
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        return result if not isInverted else 1 / result
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    def pow(self, n):
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        if n == 1:
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            return self.x
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        if n % 2 == 0:
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            p = self.pow(n / 2)
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            return p * p
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        else:
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            return self.x * self.pow(n - 1)