<p>Given <code>n</code> pairs of parentheses, write a function to <em>generate all combinations of well-formed parentheses</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 3
<strong>Output:</strong> ["((()))","(()())","(())()","()(())","()()()"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 1
<strong>Output:</strong> ["()"]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 8</code></li>
</ul>

class Solution {
public:
    void bt(vector<char>& cur, stack<char>& cur_st, vector<string>& ans, const int n, int lefts) {
        if (cur.size() == 2*n) {
            ans.push_back(string(cur.begin(), cur.end()));
            return;
        }

        if (!cur_st.empty()) {
            cur.push_back(')');
            cur_st.pop();
            bt(cur, cur_st, ans, n, lefts);
            cur_st.push('(');
            cur.pop_back();
        }
        if (lefts < n) {
            cur.push_back('(');
            cur_st.push('(');
            bt(cur, cur_st, ans, n, lefts+1);
            cur_st.pop();
            cur.pop_back();
        }

        return;

    }

    vector<string> generateParenthesis(int n) {
        vector<char> cur;
        stack<char> cur_st;
        vector<string> ans;

        bt(cur, cur_st, ans, n, 0);
        return ans;

    }
};
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