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

迷南。 2023-07-17 06:12 2阅读 0赞

> 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][Link 1]

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 ][Max Howell]的 [原问题][this original tweet] 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

##### Go 代码 #####

/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */
    func invertTree(root *TreeNode) *TreeNode { 
    	// 前序遍历
    	if root == nil { 
    		return nil
    	}
    	
    	root.Left, root.Right = root.Right, root.Left
    
    	invertTree(root.Left)
    	invertTree(root.Right)
    
    	return root
    }
    
    func invertTree(root *TreeNode) *TreeNode { 
    	// 后序遍历
    	if root == nil { 
    		return nil
    	}
    
    	invertTree(root.Left)
    	invertTree(root.Right)
    
    	root.Left, root.Right = root.Right, root.Left
    
    	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

### GitHub 链接 ###

*  [Python][]
 *  [Go][]

[LeetCode]: https://leetcode.com/problems/invert-binary-tree/
[this original tweet]: https://twitter.com/mxcl/status/608682016205344768
[Max Howell]: https://twitter.com/mxcl
[Link 1]: https://leetcode-cn.com/problems/invert-binary-tree/
[Python]: https://github.com/Wonz5130/LeetCode-Solutions/blob/master/solutions/0226-Invert-Binary-Tree/0226.py
[Go]: https://github.com/Wonz5130/LeetCode-Solutions/blob/master/solutions/0226-Invert-Binary-Tree/0226.go