101. 对称二叉树

题目描述说，最好是能用递归或迭代做

其实二叉树镜面对称，可以划分为其左子树与右子树对称
而后，比较两棵树是否对称时，相当于比较：
① A子树的左子树与B子树的右子树是否对称
② A子树的右子树与B子树的左子树是否对称
同时成立则说明两棵树对称

分析到这里，解题就比较简单了

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

class Solution:
	def isSymmetric(self, root: TreeNode) -> bool:
		def compare(tree1: TreeNode, tree2: TreeNode):
			if tree1 is None or tree2 is None:
				if tree1 is None and tree2 is None:
					return True
				else:
					return False
		
			if tree1.val != tree2.val:
				return False
			x = compare(tree1.left, tree2.right)
			y = compare(tree1.right, tree2.left)
			return x and y
		
		if root is None:
			return True
		else:
			return compare(root.left, root.right)