22.Generate Parentheses

给定一个整数n，包含 n个左括号和 n个右括号，将这n对括号组成有效的符号类型。
For example, given n = 3, a solution set is:
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]

思路来源，Grandyang：
利用递归，需要用到辅助函数，参数为 左括号left，右括号right，递归调用（left-1, right, string+'(', res）和 (left, right-1, string+')',res)，当left=right=0表示组合正常,将其结果加入容器中，而 right > left 表示右括号用的比左括号多，如 ")" 这种非法情形，就return.

class Solution {
public:
    vector<string> generateParenthrsis(int n) {
        if (n <= 0) return{};
        vector<string> res;
        generateParenthrsisDFS(n, n, "", res);
        return res;
    }
    void generateParenthrsisDFS(int left, int right, string s, vector<string>& ans) {
        if (right < left) return;
        if (left == 0 && right == 0) ans.push_back(s);
        else {
            if (left > 0) generateParenthrsisDFS(left - 1, right, s + '(', ans);
            if (right > 0) generateParenthrsisDFS(left, right - 1, s + ')', ans);
        }
    }
};

Java 版：

class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<>();
        this.generateParenthesisDFS(n, n, "", res);
        return res;
    }
    public void generateParenthesisDFS(int left, int right, String s, List<String> res){
        if(left == 0 && right == 0) res.add(s);
        else if(left > right || left < 0 || right < 0) return;
        else {
            this.generateParenthesisDFS(left - 1, right, s + "(", res);
            this.generateParenthesisDFS(left, right - 1, s + ")", res);
        }
    }
}