22. Generate Parentheses——本质：树，DFS求解可能的path

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:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

class Solution(object):
    def generateParenthesis(self, n):
        """
        :type n: int
        :rtype: List[str]
[ "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"]
   ()
   / \
 ()  ()
 / \ / \
() ()()()
        """
        ans = []
        path = []
        self.gen_par_helper(n*2, path, ans)
        return ans
    
    def is_valid_par(self, par):
        stack = []
        for c in par:
            if c == "(":
                stack.append("(")
            else:
                if stack:
                    stack.pop()
                else:
                    return False
        return len(stack) == 0
    
    def gen_par_helper(self, n, path, ans):
        if n == 0:
            if self.is_valid_par(path):
                ans.append("".join(path))
            return
        for c in "()":
            path.append(c)
            self.gen_par_helper(n-1, path, ans)
            path.pop()                        

 

class Solution(object):
    def generateParenthesis(self, n):
        """
        :type n: int
        :rtype: List[str]
[ "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"]
   ()
   / \
 ()  ()
 / \ / \
() ()()()
        """
        ans = []
        self.gen_par_helper(n, n, "", ans)
        return ans
    
    def gen_par_helper(self, left, right, path, ans):
        if left == 0 and right == 0:
            ans.append(path)
            return
        if left > 0:
            self.gen_par_helper(left-1, right, path+"(", ans)
        if right > 0 and right > left:
            self.gen_par_helper(left, right-1, path+")", ans)                        

第二种更快，只是不那么容易想到！