JZ18 二叉树镜像

题目

操作给定的二叉树，将其变换为源二叉树的镜像。

思路

• 先遍历, 节点入栈, 再依次出栈调换左右节点
• 遍历的过程中调换左右节点

代码

# -*- coding:utf-8 -*-
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 先遍历, 入栈, 再调换左右节点
    def Mirror(self, root):
        # 判断传入节点是否为空
        if root is None:
            return None
        line_node = self.printLevelNode(root)
        # print(line_node)
        while line_node:
            tmp = line_node.pop()
            if tmp.left or tmp.right:
                tmp.left, tmp.right = tmp.right, tmp.left
        return tmp
    # 层次遍历二叉树, 被调用
    def printLevelNode(self, root):
        line_node = []
        res = []
        line_node.append(root)
        while line_node:
            tmp = line_node.pop(0)
            res.append(tmp)
            if tmp.left:
                line_node.append(tmp.left)
            if tmp.right:
                line_node.append(tmp.right)
        return res

    # 层次遍历的过程中调换左右节点
    def Mirror2(self, root):
        if root is None:
            return None
        line_node = []
        line_node.append(root)
        while line_node:
            tmp = line_node.pop(0)
            if tmp.left:
                line_node.append(tmp.left)
            if tmp.right:
                line_node.append(tmp.right)
            # if tmp.left or tmp.right:
            tmp.left, tmp.right = tmp.right, tmp.left
        return root

    # 递归遍历的过程中调换左右节点
    def Mirror3(self, root):
        if root is None:
            return None
        root.left, root.right = root.right, root.left
        self.Mirror3(root.left)
        self.Mirror3(root.right)
        return root


if __name__ == '__main__':
    node1 = TreeNode(8)
    node2 = TreeNode(6)
    node3 = TreeNode(10)
    node4 = TreeNode(5)
    node5 = TreeNode(7)
    node6 = TreeNode(9)
    node7 = TreeNode(11)
    node1.left = node2
    node1.right = node3
    node2.left = node4
    node2.right = node5
    node3.left = node6
    node3.right = node7
    sl = Solution()
    ls = sl.Mirror3(node1)
    print(ls)
    for i in sl.printLevelNode(ls):
        print(i.val)