题目：

解法：

The idea is intuitive. Use two integers to count the remaining left parenthesis (n) and the right parenthesis (m) to be added. At each function call add a left parenthesis if n >0 and add a right parenthesis if m>0. Append the result and terminate recursive calls when both m and n are zero.

class Solution 
{
public:vector<string> generateParenthesis(int n) {vector<string> res;addingpar(res, "", n, 0);return res;}void addingpar(vector<string> &v, string str, int n, int m){if(n==0 && m==0) {v.push_back(str);return;}if(m > 0){ addingpar(v, str+")", n, m-1); }if(n > 0){ addingpar(v, str+"(", n-1, m+1); }}
};