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{
"data": {
"question": {
"questionId": "99",
"questionFrontendId": "99",
"categoryTitle": "Algorithms",
"boundTopicId": 1173,
"title": "Recover Binary Search Tree",
"titleSlug": "recover-binary-search-tree",
"content": "<p>You are given the <code>root</code> of a binary search tree (BST), where the values of <strong>exactly</strong> two nodes of the tree were swapped by mistake. <em>Recover the tree without changing its structure</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/28/recover1.jpg\" style=\"width: 422px; height: 302px;\" />\n<pre>\n<strong>Input:</strong> root = [1,3,null,null,2]\n<strong>Output:</strong> [3,1,null,null,2]\n<strong>Explanation:</strong> 3 cannot be a left child of 1 because 3 &gt; 1. Swapping 1 and 3 makes the BST valid.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/28/recover2.jpg\" style=\"width: 581px; height: 302px;\" />\n<pre>\n<strong>Input:</strong> root = [3,1,4,null,null,2]\n<strong>Output:</strong> [2,1,4,null,null,3]\n<strong>Explanation:</strong> 2 cannot be in the right subtree of 3 because 2 &lt; 3. Swapping 2 and 3 makes the BST valid.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the tree is in the range <code>[2, 1000]</code>.</li>\n\t<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</code></li>\n</ul>\n\n<p>&nbsp;</p>\n<strong>Follow up:</strong> A solution using <code>O(n)</code> space is pretty straight-forward. Could you devise a constant <code>O(1)</code> space solution?",
"translatedTitle": "恢复二叉搜索树",
"translatedContent": "<p>给你二叉搜索树的根节点 <code>root</code> ,该树中的 <strong>恰好</strong> 两个节点的值被错误地交换。<em>请在不改变其结构的情况下,恢复这棵树&nbsp;</em>。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/28/recover1.jpg\" style=\"width: 300px;\" />\n<pre>\n<strong>输入:</strong>root = [1,3,null,null,2]\n<strong>输出:</strong>[3,1,null,null,2]\n<strong>解释:</strong>3 不能是 1 的左孩子,因为 3 &gt; 1 。交换 1 和 3 使二叉搜索树有效。\n</pre>\n\n<p><strong>示例 2</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/28/recover2.jpg\" style=\"height: 208px; width: 400px;\" />\n<pre>\n<strong>输入:</strong>root = [3,1,4,null,null,2]\n<strong>输出:</strong>[2,1,4,null,null,3]\n<strong>解释:</strong>2 不能在 3 的右子树中,因为 2 &lt; 3 。交换 2 和 3 使二叉搜索树有效。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>树上节点的数目在范围 <code>[2, 1000]</code> 内</li>\n\t<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶:</strong>使用 <code>O(n)</code> 空间复杂度的解法很容易实现。你能想出一个只使用&nbsp;<code>O(1)</code> 空间的解决方案吗?</p>\n",
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"difficulty": "Medium",
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"topicTags": [
{
"name": "Tree",
"slug": "tree",
"translatedName": "树",
"__typename": "TopicTagNode"
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{
"name": "Depth-First Search",
"slug": "depth-first-search",
"translatedName": "深度优先搜索",
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{
"name": "Binary Search Tree",
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"translatedName": "二叉搜索树",
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"code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * TreeNode *left;\n * TreeNode *right;\n * TreeNode() : val(0), left(nullptr), right(nullptr) {}\n * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}\n * };\n */\nclass Solution {\npublic:\n void recoverTree(TreeNode* root) {\n\n }\n};",
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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() {}\n * TreeNode(int val) { this.val = val; }\n * TreeNode(int val, TreeNode left, TreeNode right) {\n * this.val = val;\n * this.left = left;\n * this.right = right;\n * }\n * }\n */\nclass Solution {\n public void recoverTree(TreeNode root) {\n\n }\n}",
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"lang": "C",
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"code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * struct TreeNode *left;\n * struct TreeNode *right;\n * };\n */\nvoid recoverTree(struct TreeNode* root) {\n \n}",
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"code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n * public int val;\n * public TreeNode left;\n * public TreeNode right;\n * public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {\n * this.val = val;\n * this.left = left;\n * this.right = right;\n * }\n * }\n */\npublic class Solution {\n public void RecoverTree(TreeNode root) {\n\n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * function TreeNode(val, left, right) {\n * this.val = (val===undefined ? 0 : val)\n * this.left = (left===undefined ? null : left)\n * this.right = (right===undefined ? null : right)\n * }\n */\n/**\n * @param {TreeNode} root\n * @return {void} Do not return anything, modify root in-place instead.\n */\nvar recoverTree = function(root) {\n\n};",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * val: number\n * left: TreeNode | null\n * right: TreeNode | null\n * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {\n * this.val = (val===undefined ? 0 : val)\n * this.left = (left===undefined ? null : left)\n * this.right = (right===undefined ? null : right)\n * }\n * }\n */\n\n/**\n Do not return anything, modify root in-place instead.\n */\nfunction recoverTree(root: TreeNode | null): void {\n \n};",
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"code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n * public var val: Int\n * public var left: TreeNode?\n * public var right: TreeNode?\n * public init() { self.val = 0; self.left = nil; self.right = nil; }\n * public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }\n * public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {\n * self.val = val\n * self.left = left\n * self.right = right\n * }\n * }\n */\nclass Solution {\n func recoverTree(_ root: TreeNode?) {\n\n }\n}",
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"lang": "Kotlin",
"langSlug": "kotlin",
"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 recoverTree(root: TreeNode?): Unit {\n\n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * int val;\n * TreeNode? left;\n * TreeNode? right;\n * TreeNode([this.val = 0, this.left, this.right]);\n * }\n */\nclass Solution {\n void recoverTree(TreeNode? root) {\n \n }\n}",
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"lang": "Go",
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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 recoverTree(root *TreeNode) {\n\n}",
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"lang": "Racket",
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