leetcode 之 Symmetric Tree

题目描述：
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 
    1
   / 
  2   2
 /  / 
3  4 4  3



But the following [1,2,2,null,3,null,3] is not:

    1
   / 
  2   2
      
   3    3

思路：题目的要求为判断一棵树是否对称， 可以从根节点向下，将二叉树看成两颗树， 判断这两棵树是否对称。采用递归的方式，代码如下：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        nums = []
        st = []
        if not root:
            return True
        return self.helper(root.left, root.right)
        
    def helper(self, left, right):
        if not left and not right:
            return True
        elif not left or not right:
            return False
        if left.val != right.val:
            return False
        return self.helper(left.left, right.right) and self.helper(left.right, right.left)

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = None
class Solution(object):    def isSymmetric(self, root):        """        :type root: TreeNode        :rtype: bool        """        nums = []        st = []        if not root:            return True        return self.helper(root.left, root.right)            def helper(self, left, right):        if not left and not right:            return True        elif not left or not right:            return False        if left.val != right.val:            return False        return self.helper(left.left, right.right) and self.helper(left.right, right.left)