class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        dfs(0,0, "", n,res);
        return res;
    }
    void dfs(int openP, int closeP, string s, int n, vector<string> & res){
        if(openP == closeP && openP+closeP == 2*n){
            res.push_back(s);
            return;
        }

        if(openP<n){
            dfs(openP+1, closeP, s+'(', n ,res);
        }
        if(closeP<openP){
            dfs(openP, closeP+1, s+')',n,res);
        }
    }
};

//The recursion goes as deep as possible by completing one full parentheses string before coming back and trying another option.
//Each recursive call explores one path fully, then backtracks — which is exactly depth-first search
//Note we are passing s by value, so no need to reset it for every path