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
Note: Bonus points if you could solve it both recursively and iteratively.
解题思路：
简单方法，就是复制一棵相同的数，然后比较左右结点是否满足对称树的条件。由于原函数只传入了一棵树的根节点，因此需要重新定义一个函数，可以传入两棵树的根节点，然后进行比较。
Python实现：
# 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):
"""
:type root: TreeNode
:rtype: bool
"""
if root == None:
return True
return Solution.isMirror(self, root.left, root.right)
def isMirror(self, p, q):
if p == None and q == None:
return True
if p != None and q != None and p.val == q.val:
return Solution.isMirror(self, p.left, q.right) and Solution.isMirror(self, p.right, q.left)
else:
return False
