LF.66.All Valid Permutations Of Parentheses I

same as LC.22. Generate Parentheses
https://leetcode.com/problems/generate-parentheses/description/

Given N pairs of parentheses “()”, return a list with all the valid permutations.

Assumptions

N >= 0
Examples

N = 1, all valid permutations are ["()"]
N = 3, all valid permutations are ["((()))", "(()())", "(())()", "()(())", "()()()"]
N = 0, all valid permutations are [""]

public class Solution {
  public List<String> validParentheses(int n) {
    // Write your solution here.
    //assume n>=0
    List<String> res = new ArrayList<>() ;
    StringBuilder sol = new StringBuilder() ;
    dfs(n, res, 0, 0, sol) ;
    return res ;
  }

 /*
    n stores total number of "pair of ()" need to add.
    So total levels == 2*n
    1 stores the number of left parenthesis "(" added so far
    r stores the number of right parenthesis ")" added so far
    sol: solution so far
 */
  private void dfs(int n, List<String> res, int l , int r , StringBuilder sol){
     //base case: reach the leaf and its valid value, go back
    if (n == l && l == r ) {
        //always deep copy, be very careful
        res.add(sol.toString());
        return ;
    }
    //case 1: add ( on this level :
    //think n as the length of array, l as index
    if (n > l) {
        dfs(n, res, l+1 , r , sol.append("(")) ;
        //remove
        sol.deleteCharAt(sol.length()-1) ;
    }
    //case 2: add )
    if (n > r && l > r ) {
        dfs(n, res, l, r+1, sol.append(")")) ;
        //remove
        sol.deleteCharAt(sol.length()-1) ;
    }
  }
}