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leetcode-problemset/leetcode-cn/originData/er-cha-shu-de-jing-xiang-lcof.json
2023-12-09 19:57:46 +08:00

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{
"data": {
"question": {
"questionId": "100288",
"questionFrontendId": "LCR 144",
"categoryTitle": "Algorithms",
"boundTopicId": 86201,
"title": "翻转二叉树",
"titleSlug": "er-cha-shu-de-jing-xiang-lcof",
"content": "English description is not available for the problem. Please switch to Chinese.",
"translatedTitle": "翻转二叉树",
"translatedContent": "<p>给定一棵二叉树的根节点 <code>root</code>,请左右翻转这棵二叉树,并返回其根节点。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<p><img alt=\"\" src=\"https://pic.leetcode.cn/1694686821-qlvjod-%E7%BF%BB%E8%BD%AC%E4%BA%8C%E5%8F%89%E6%A0%91.png\" /></p>\n\n<pre>\n<strong>输入:</strong>root = [5,7,9,8,3,2,4]\n<strong>输出:</strong>[5,9,7,4,2,3,8]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>树中节点数目范围在 <code>[0, 100]</code> 内</li>\n\t<li><code>-100 &lt;= Node.val &lt;= 100</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p>注意:本题与主站 226 题相同:<a href=\"https://leetcode-cn.com/problems/invert-binary-tree/\">https://leetcode-cn.com/problems/invert-binary-tree/</a></p>\n\n<p>&nbsp;</p>\n",
"isPaidOnly": false,
"difficulty": "Easy",
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"name": "Binary Tree",
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"code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n * int val;\n * TreeNode left;\n * TreeNode right;\n * TreeNode(int x) { val = x; }\n * }\n */\nclass Solution {\n public TreeNode mirrorTree(TreeNode root) {\n\n }\n}",
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"code": "# Definition for a binary tree node.\n# class TreeNode(object):\n# def __init__(self, x):\n# self.val = x\n# self.left = None\n# self.right = None\n\nclass Solution(object):\n def mirrorTree(self, root):\n \"\"\"\n :type root: TreeNode\n :rtype: TreeNode\n \"\"\"",
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"code": "# Definition for a binary tree node.\n# class TreeNode:\n# def __init__(self, x):\n# self.val = x\n# self.left = None\n# self.right = None\n\nclass Solution:\n def mirrorTree(self, root: TreeNode) -> TreeNode:",
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"code": "/**\n * Definition for a binary tree node.\n * function TreeNode(val) {\n * this.val = val;\n * this.left = this.right = null;\n * }\n */\n/**\n * @param {TreeNode} root\n * @return {TreeNode}\n */\nvar mirrorTree = function(root) {\n\n};",
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"code": "/**\n * Example:\n * var ti = TreeNode(5)\n * var v = ti.`val`\n * Definition for a binary tree node.\n * class TreeNode(var `val`: Int) {\n * var left: TreeNode? = null\n * var right: TreeNode? = null\n * }\n */\nclass Solution {\n fun mirrorTree(root: TreeNode?): TreeNode? {\n\n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * type TreeNode struct {\n * Val int\n * Left *TreeNode\n * Right *TreeNode\n * }\n */\nfunc mirrorTree(root *TreeNode) *TreeNode {\n\n}",
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"code": "// Definition for a binary tree node.\n// #[derive(Debug, PartialEq, Eq)]\n// pub struct TreeNode {\n// pub val: i32,\n// pub left: Option<Rc<RefCell<TreeNode>>>,\n// pub right: Option<Rc<RefCell<TreeNode>>>,\n// }\n// \n// impl TreeNode {\n// #[inline]\n// pub fn new(val: i32) -> Self {\n// TreeNode {\n// val,\n// left: None,\n// right: None\n// }\n// }\n// }\nuse std::rc::Rc;\nuse std::cell::RefCell;\nimpl Solution {\n pub fn mirror_tree(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {\n\n }\n}",
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{
"lang": "Racket",
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"code": "; Definition for a binary tree node.\n#|\n\n; val : integer?\n; left : (or/c tree-node? #f)\n; right : (or/c tree-node? #f)\n(struct tree-node\n (val left right) #:mutable #:transparent)\n\n; constructor\n(define (make-tree-node [val 0])\n (tree-node val #f #f))\n\n|#\n\n(define/contract (mirror-tree root)\n (-> (or/c tree-node? #f) (or/c tree-node? #f))\n\n )",
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"code": "# Definition for a binary tree node.\n#\n# defmodule TreeNode do\n# @type t :: %__MODULE__{\n# val: integer,\n# left: TreeNode.t() | nil,\n# right: TreeNode.t() | nil\n# }\n# defstruct val: 0, left: nil, right: nil\n# end\n\ndefmodule Solution do\n @spec mirror_tree(root :: TreeNode.t | nil) :: TreeNode.t | nil\n def mirror_tree(root) do\n\n end\nend",
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