二叉树的镜像和对称的二叉树

二叉树的镜像

题目描述：
操作给定的二叉树，将其变换为源二叉树的镜像。

代码实现：

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 返回镜像树的根节点
    def Mirror(self, root):
        # write code here
        if root:
            root.left,root.right = root.right,root.left
            self.Mirror(root.left)
            self.Mirror(root.right)

对称的二叉树

题目描述：
请实现一个函数，用来判断一颗二叉树是不是对称的。注意，如果一个二叉树同此二叉树的镜像是同样的，定义其为对称的。
代码实现：

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def fun(self,node1,node2):
        if not node1 and not node2:
            return True
        if node1 and node2 and node1.val == node2.val:
            return self.fun(node1.left, node2.right) and self.fun(node1.right, node2.left)
        return False
    def isSymmetrical(self, pRoot):
        # write code here
        if not pRoot:
            return True
        return self.fun(pRoot.left,pRoot.right)

【补充】可以用对称的二叉树思路判断两个二叉树是否完全一致

def EqTree(self,TreeNode1,TreeNode2):
        if not TreeNode1 and not TreeNode2:
            return Ture
        if TreeNode1 and TreeNode2 and TreeNode1.val == TreeNode2.val:
            return Tree(TreeNode1.left,TreeNode2.left) and  Tree(TreeNode1.right,TreeNode2.right) 
        else:
            return False