leetcode 101. Symmetric Tree

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

思路：对称树的结构是递归的，我们也用递归去遍历节点。因为根节点与其他节点有些不同，所以需要再定义一个函数去遍历节点，而初始给的函数来作为开始。初始只需要判断根节点是否为空，是就return True,否就调用我们定义的函数。然后每次都先判断是否都为空？是否一空一不空？都不空的话值是否相等，若是相等，就要用inpair去遍历整个树结构中靠中间的部分，outpair去遍历整个树结构中靠两边的部分，再返回判断出的布尔值；若不等就直接返回False了。这样最后返回的就是一串布尔值的and运算。

class Solution(object):
def isSymmetric(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if root == None:
return True
else:
return self.isMirror(root.left, root.right)

def isMirror(self, left, right):
if left == None and right == None:
return True
if left == None or right == None:
return False
if left.val == right.val:
inpair = self.isMirror(left.right, right.left)
outpair = self.isMirror(left.left, right.right)
return inpair and outpair
else:
return False