剑指 Offer 28. 对称的二叉树

      1. not left and right      2. left and not right      3. left.val!=right.val

# 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:        '''            方法：广度优先搜索        '''        if not root:            return True         return self.bfs(root.left,root.right)
    # 方法：广度优先搜索    def bfs(self,left,right):        # step 1：left 和 right 都为空，        #         也就是父节点为叶子节点且相同        if not left and not right:            return True         # step 2：三种非对称树的情况：        #    1. not left and right        #    2. left and not right        #    3. left.val!=right.val        elif (not left and right) or (left and not right) or (left.val!=right.val):            return False
        return self.bfs(left.left,right.right) and self.bfs(left.right,right.left)