184 lines
3.7 KiB
Markdown
184 lines
3.7 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/invert-binary-tree/description/
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## 题目描述
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```
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Invert a binary tree.
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Example:
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Input:
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4
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/ \
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2 7
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/ \ / \
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1 3 6 9
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Output:
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4
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/ \
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7 2
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/ \ / \
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9 6 3 1
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Trivia:
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This problem was inspired by this original tweet by Max Howell:
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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.
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```
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## 思路
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遍历树（随便怎么遍历），然后将左右子树交换位置。
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## 关键点解析
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- 递归简化操作
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- 如果树很高，建议使用栈来代替递归
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- 这道题目对顺序没要求的，因此队列数组操作都是一样的，无任何区别
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## 代码
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* 语言支持：JS，Python，C++
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Javascript Code:
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```js
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/*
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* @lc app=leetcode id=226 lang=javascript
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*
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* [226] Invert Binary Tree
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*
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* https://leetcode.com/problems/invert-binary-tree/description/
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*
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* algorithms
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* Easy (57.14%)
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* Total Accepted: 311K
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* Total Submissions: 540.6K
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* Testcase Example: '[4,2,7,1,3,6,9]'
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*
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* Invert a binary tree.
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*
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* Example:
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*
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* Input:
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*
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*
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* 4
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* / \
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* 2 7
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* / \ / \
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* 1 3 6 9
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*
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* Output:
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*
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*
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* 4
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* / \
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* 7 2
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* / \ / \
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* 9 6 3 1
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*
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* Trivia:
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* This problem was inspired by this original tweet by Max Howell:
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*
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* Google: 90% of our engineers use the software you wrote (Homebrew), but you
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* can’t invert a binary tree on a whiteboard so f*** off.
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*
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*/
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/**
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* Definition for a binary tree node.
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* function TreeNode(val) {
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* this.val = val;
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* this.left = this.right = null;
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* }
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*/
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/**
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* @param {TreeNode} root
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* @return {TreeNode}
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*/
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var invertTree = function(root) {
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if (!root) return root;
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// 递归
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// const left = root.left;
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// const right = root.right;
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// root.right = invertTree(left);
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// root.left = invertTree(right);
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// 我们用stack来模拟递归
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// 本质上递归是利用了执行栈，执行栈也是一种栈
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// 其实这里使用队列也是一样的，因为这里顺序不重要
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const stack = [root];
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let current = null;
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while ((current = stack.shift())) {
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const left = current.left;
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const right = current.right;
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current.right = left;
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current.left = right;
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if (left) {
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stack.push(left);
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}
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if (right) {
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stack.push(right);
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}
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}
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return root;
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};
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```
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Python Code:
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def invertTree(self, root: TreeNode) -> TreeNode:
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if not root:
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return None
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stack = [root]
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while stack:
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node = stack.pop(0)
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node.left, node.right = node.right, node.left
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if node.left:
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stack.append(node.left)
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if node.right:
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stack.append(node.right)
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return root
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```
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C++ Code：
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```C++
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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TreeNode* invertTree(TreeNode* root) {
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if (root == NULL) return root;
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auto q = queue<TreeNode*>();
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q.push(root);
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while (!q.empty()) {
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auto n = q.front(); q.pop();
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swap(n->left, n->right);
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if (n->left != nullptr) {
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q.push(n->left);
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}
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if (n->right != nullptr) {
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q.push(n->right);
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}
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}
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return root;
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}
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};
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```
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