[leetcode:python]101. Symmetric Tree

题目：
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

方法一：性能52ms

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def help(self, p, q):        if p == None and q == None: return True        if p and q and p.val == q.val:            return self.help(p.right, q.left) and self.help(p.left, q.right)        return False    def isSymmetric(self, root):        """        :type root: TreeNode        :rtype: bool        """        if root:            return self.help(root.left, root.right)        return True

方法二：性能38ms

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def isSymmetric(self, root):        """        :type root: TreeNode        :rtype: bool        """        if root is None:            return True        return self.isSymmetricNodes(root.left, root.right)    def isSymmetricNodes(self, l, r):        if l is None and r is None:            return True        elif l is None or r is None:            return False        return (l.val == r.val) and  self.isSymmetricNodes(l.left, r.right) and self.isSymmetricNodes(l.right, r.left)