Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
class Solution(object):
def isSymmetric(self, root):
if not root:
return True
def dfs(left,right):
# 递归的终止条件是两个节点都为空
# 或者两个节点中有一个为空
# 或者两个节点的值不相等
if not (left or right):
return True
if not (left and right):
return False
if left.val!=right.val:
return False
return dfs(left.left,right.right) and dfs(left.right,right.left)
# 用递归函数,比较左节点,右节点
return dfs(root.left,root.right)
标签:right,return,Tree,dfs,Symmetric,root,节点,left
From: https://www.cnblogs.com/artwalker/p/17513431.html