LeetCode | 0226. Invert Binary Tree翻转二叉树【Python】

LeetCode 0226. Invert Binary Tree翻转二叉树【Easy】【Python】【二叉树】【递归】

Problem

LeetCode

Invert a binary tree.

Example:

Input:

     4
   /   \
  2     7
 / \   / \
1   3 6   9

Output:

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

问题

力扣

翻转一棵二叉树。

示例：

输入：

     4
   /   \
  2     7
 / \   / \
1   3 6   9

输出：

     4
   /   \
  7     2
 / \   / \
9   6 3   1

备注:
这个问题是受到 Max Howell 的 原问题 启发的 ：

谷歌：我们90％的工程师使用您编写的软件(Homebrew)，但是您却无法在面试时在白板上写出翻转二叉树这道题，这太糟糕了。

思路

解法一

递归

前序遍历二叉树，如果当前节点有子树，就交换左右子树。

Python3代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        # solution one: 递归
        if not root:
            return None
        # 叶子节点，直接返回自己
        if not root.left and not root.right:
            return root
        
        # 交换非叶子节点的左右两棵子树
        root.left, root.right = root.right, root.left
        if root.left:
            self.invertTree(root.left)
        if root.right:
            self.invertTree(root.right)
        return root

解法二

栈

用栈模拟二叉树。

Python3代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        # solution two: 栈
        if not root:
            return None
        # 叶子节点，直接返回自己
        if not root.left and not root.right:
            return root
        
        # 栈模拟二叉树
        stack = [root]
        while stack:
            node = stack.pop()
            if node:
                node.left, node.right = node.right, node.left
                stack.append(node.right)
                stack.append(node.left)
        return root

代码地址

GitHub链接