题目思路：

给出 n 代表生成括号的对数，请你写出一个函数，使其能够生成所有可能的并且有效的括号组合。

例如，给出 n = 3，生成结果为：

[

    "((()))",

    "(()())",

    "(())()",

    "()(())",

    "()()()"

]

class Solution(object):
    def generateParenthesis(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        #思路：定义一个括号生成的递归函数，第一个参数为有效的括号组合，第二个参数为当前的括号字符串，第三个参数为括号的对数，第四个参数为当前括号字符串中左括号的个数，第五个参数为当前括号字符串中右括号的个数
        res = []
        temp = '('
        l_num = 1
        r_num = 0
        
        if n == 0:
            return res
        
        def generate(list_,ss,nums,l_nums,r_nums):
            if nums == l_nums and nums == r_nums:
                list_.append(ss)
            elif nums == l_nums and nums != r_nums:
                ss = ss + (nums-r_nums)*')'
                list_.append(ss)
            else:
                if l_nums == r_nums:
                    ss = ss + '('
                    generate(list_,ss,nums,l_nums+1,r_nums)
                else:
                    tt = ss + '('
                    generate(list_,tt,nums,l_nums+1,r_nums)
                    tt = ss + ')'
                    generate(list_,tt,nums,l_nums,r_nums+1)
                    
        generate(res,temp,n,l_num,r_num)
        
        return res

菜鸟一枚，代码仅供参考，如有问题，望指正~