括号生成，力扣上原题，有一些疑惑

Question

class Solution {
void backtrack(vector& ans, string& cur, int open, int close, int n) {
if (cur.size() == n * 2) {
ans.push_back(cur);
return;
}
if (open < n) {
cur.push_back('(');
backtrack(ans, cur, open + 1, close, n);
cur.pop_back();
}
if (close < open) {
cur.push_back(')');
backtrack(ans, cur, open, close + 1, n);
cur.pop_back();
}
}
public:
vector generateParenthesis(int n) {
vector result;
string current;
backtrack(result, current, 0, 0, n);
return result;
}
};
问：来一个认真负责的大佬，详细列举举例一下回溯的过程，我一直不停的推导void backtrack，例如n=3，我只得到了（（（）））一个结果，希望详细一点
力扣上的原题的官方解析，回溯法，来一个认真负责的大佬

Answer 1

#include <vector>
#include <string>
#include <iostream>
using namespace std;

class Solution
{
    void backtrack(vector<string> &ans, string &cur, int open, int close, int n)
    {
        if (cur.size() == n * 2)
        {
            ans.push_back(cur);
            return;
        }
        if (open < n)
        {
            cur.push_back('(');
            cout << cur << endl;
            backtrack(ans, cur, open + 1, close, n);
            cur.pop_back();
            cout << cur << endl;
        }
        if (close < open)
        {
            cur.push_back(')');
            cout << cur << endl;
            backtrack(ans, cur, open, close + 1, n);
            cur.pop_back();
            cout << cur << endl;
        }
    }

public:
    vector<string> generateParenthesis(int n)
    {
        vector<string> result;
        string current;
        backtrack(result, current, 0, 0, n);
        return result;
    }
};

int main(void)
{
    Solution su;
    vector<string> r = su.generateParenthesis(2);
    cout << "result is: " << endl;
    for (int i = 0; i < r.size(); ++i)
    {
        cout << r[i] << endl;
    }
    return 0;
}

打印每个步骤看一下：
以n=2为例子:
1.
current = ""
backtrack(result, current, 0, 0, 2);

2.
open = 0, close = 0
current = "("
backtrack(result, current, 1, 0, 2); --> 进入if(open < n)内的backtrack

3.
open = 1, close = 0
current = "(("
backtrack(result, current, 2, 0, 2); --> 进入if(open < n)内的backtrack

4.
open = 2, close = 0
current = "(()"
backtrack(ans, cur, 2, 1, 2); --> 进入if(close < open)内的backtrack

5.
open = 2, close = 1
current = "(())"
backtrack(ans, cur, 2, 2, 2); --> 进入if(close < open)内的backtrack

6.
open = 2, close = 2
此时current.size() == 4
执行 ans.push_back(cur); --> 得到第一个合格字符串"(())"

1. 继续执行第5步

   if (close < open) {
   cur.push_back(')');
   backtrack(ans, cur, open, close + 1, n);
   cur.pop_back();
   }
   

   current = "(()"

2. 继续执行第4步

   if (close < open) {
   cur.push_back(')');
   backtrack(ans, cur, open, close + 1, n);
   cur.pop_back();
   }
   

   current = "(("

3. 继续执行第3步

   if (open < n) {
   cur.push_back('(');
   backtrack(ans, cur, open + 1, close, n);
   cur.pop_back();
   }
   

   current = "("

并且第3步中: open = 1, close = 0, close < open
执行

cur.push_back(')');
backtrack(ans, cur, open, close + 1, n);

current = "()"
backtrack(ans, cur, 1, 1, 2) --> 进入if(close < open)内的backtrack

10.
open = 1, close = 1
current = "()("
backtrack(ans, cur, 2, 1, 2); --> 进入if(open < n)内的backtrack

11.
open = 2, close = 1
current = "()()"
backtrack(ans, cur, 2, 2, 2) --> 进入if(close < open)内的backtrack

13.
open = 2, close = 2
此时current.size() == 4
执行 ans.push_back(cur); --> 得到第二个合格字符串"()()"