leetcode 22. Generate Parentheses-回溯|递归

原题链接：22. Generate Parentheses

【思路-Java、Python】 递归实现

采用递归方式，添加左括号的条件是左括号数量小于 n，添加右括号的条件是右括号数小于左括号数：

public class Solution {
    public List generateParenthesis(int n) {
        List res = new ArrayList();
        dfs(res, "", n, 0, 0);
        return res;
    }
    public void dfs(List res, String temp, int n, int left, int right) {
        if(right == n) {
            res.add(temp);
            return;
        }
        if (left < n) dfs(res, temp+"(", n, left+1, right);
        if (right < left) dfs(res, temp+")", n, left, right+1);
    }
}

8 / 8  test cases passed. Runtime: 2 ms  Your runtime beats 26.33% of javasubmissions.

class Solution(object):
    def generateParenthesis(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        res = []
        def dfs(res, n, s, left, right) :
            if right == n : res.append(s)
            if left < n :
                dfs(res, n, s + "(", left+1, right)
            if left > right :
                dfs(res, n, s + ")", left, right+1)
        dfs(res, n, '', 0, 0)
        return res

8 / 8  test cases passed. 8 / 8 test cases passed.  Your runtime beats 66.46% of pythonsubmissions.