剑指offer 二叉树的镜像（二叉树的所有子节点互换）_给一个二叉树,子节点互换-优快云博客

本文链接：https://blog.youkuaiyun.com/lijiaqi0612/article/details/101068141

本文介绍了一种将二叉树变换为其镜像的算法，提供了递归和非递归两种实现方式，通过实例详细解释了二叉树及其镜像的概念，并给出了具体的代码实现。

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题目描述

操作给定的二叉树，将其变换为源二叉树的镜像。

输入描述:

二叉树的镜像定义：源二叉树 
    	    8
    	   /  \
    	  6   10
    	 / \  / \
    	5  7 9 11
    	镜像二叉树
    	    8
    	   /  \
    	  10   6
    	 / \  / \
    	11 9 7  5

解法一：递归

# -*- 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 == None:
            return 
        root.left, root.right = root.right, root.left
        if root.left:
            self.Mirror(root.left)
        if root.right:
            self.Mirror(root.right)

解法二：非递归

# -*- 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 == None:
            return 


        stackNode = []
        stackNode.append(root)
        while len(stackNode) > 0:
            nodeNum = len(stackNode) - 1
            tree = stackNode[nodeNum]
            stackNode.pop()
            nodeNum -= 1
            if tree.left != None or tree.right != None:
                tree.left, tree.right = tree.right, tree.left
            if tree.left:
                stackNode.append(tree.left)
                nodeNum += 1
            if tree.right:
                stackNode.append(tree.right)
                nodeNum += 1