Generate Parentheses

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"

Analysis: DFS. The key point is that the number of right parentheses cannot greater than the number of left parentheses. 

public class Solution {
public void generateParenthesis(int n, int left, int right, ArrayList<String> res, StringBuilder tem) {
if(left==n && right==n) {
res.add(tem.toString());
return;
}

if(left < n) {
tem.append("(");
generateParenthesis(n, left+1, right, res, tem);
tem.deleteCharAt(tem.length()-1);
}
if(right<n && right+1<=left) {
tem.append(")");
generateParenthesis(n, left, right+1, res, tem);
tem.deleteCharAt(tem.length()-1);
}
return;
}

public ArrayList<String> generateParenthesis(int n) {
ArrayList<String> res = new ArrayList<String>();
StringBuilder tem = new StringBuilder();
generateParenthesis(n, 0, 0, res, tem);
return res;
}
}