对称二叉树

算法1

class Solution(object):
def isSymmetric(self, root):
“”“
:type root: TreeNode
:rtype: bool
“”“
if not root:
return True

    return self.isSym(root.left,root.right)

def isSym(self,left,right):
    if not left or not right:
        return not left and not right
    if left.val != right.val:
        return False
    return self.isSym(left.left,right.right) and self.isSym(left.right,right.left)

参考y总的算法写的，那个not left or not right判断条件写得太棒了