Leetcode 22

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Question

Solution

Dfs来做，考虑left和right的各自的次数，每添加一个就-1，然后通过判断语句来剔除不正确的情况。

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class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        if n == 0:
            return [""]
        results = []
        self.dfs(n, n, "", results)
        return results
    
    
    def dfs(self, left, right, temp, results):
        if left > right or left < 0 or right < 0:
            return 
        if left == 0 and right == 0:
            results.append(temp)
        self.dfs(left - 1, right, temp + "(", results)
        self.dfs(left, right - 1, temp + ")", results)