{ "data": { "question": { "questionId": "2060", "questionFrontendId": "1932", "categoryTitle": "Algorithms", "boundTopicId": 863111, "title": "Merge BSTs to Create Single BST", "titleSlug": "merge-bsts-to-create-single-bst", "content": "

You are given n BST (binary search tree) root nodes for n separate BSTs stored in an array trees (0-indexed). Each BST in trees has at most 3 nodes, and no two roots have the same value. In one operation, you can:

\n\n\n\n

Return the root of the resulting BST if it is possible to form a valid BST after performing n - 1 operations, or null if it is impossible to create a valid BST.

\n\n

A BST (binary search tree) is a binary tree where each node satisfies the following property:

\n\n\n\n

A leaf is a node that has no children.

\n\n

 

\n

Example 1:

\n\"\"\n
\nInput: trees = [[2,1],[3,2,5],[5,4]]\nOutput: [3,2,5,1,null,4]\nExplanation:\nIn the first operation, pick i=1 and j=0, and merge trees[0] into trees[1].\nDelete trees[0], so trees = [[3,2,5,1],[5,4]].\n\"\"\nIn the second operation, pick i=0 and j=1, and merge trees[1] into trees[0].\nDelete trees[1], so trees = [[3,2,5,1,null,4]].\n\"\"\nThe resulting tree, shown above, is a valid BST, so return its root.
\n\n

Example 2:

\n\"\"\n
\nInput: trees = [[5,3,8],[3,2,6]]\nOutput: []\nExplanation:\nPick i=0 and j=1 and merge trees[1] into trees[0].\nDelete trees[1], so trees = [[5,3,8,2,6]].\n\"\"\nThe resulting tree is shown above. This is the only valid operation that can be performed, but the resulting tree is not a valid BST, so return null.\n
\n\n

Example 3:

\n\"\"\n
\nInput: trees = [[5,4],[3]]\nOutput: []\nExplanation: It is impossible to perform any operations.\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "合并多棵二叉搜索树", "translatedContent": "

给你 n 个 二叉搜索树的根节点 ，存储在数组 trees 中（下标从 0 开始），对应 n 棵不同的二叉搜索树。trees 中的每棵二叉搜索树 最多有 3 个节点 ，且不存在值相同的两个根节点。在一步操作中，将会完成下述步骤：

\n\n\n\n

如果在执行 n - 1 次操作后，能形成一棵有效的二叉搜索树，则返回结果二叉树的 根节点 ；如果无法构造一棵有效的二叉搜索树，返回 null 。

\n\n

二叉搜索树是一种二叉树，且树中每个节点均满足下述属性：

\n\n\n\n

叶节点是不含子节点的节点。

\n\n

 

\n\n

示例 1：

\n\"\"\n
\n输入：trees = [[2,1],[3,2,5],[5,4]]\n输出：[3,2,5,1,null,4]\n解释：\n第一步操作中，选出 i=1 和 j=0 ，并将 trees[0] 合并到 trees[1] 中。\n删除 trees[0] ，trees = [[3,2,5,1],[5,4]] 。\n\"\"\n在第二步操作中，选出 i=0 和 j=1 ，将 trees[1] 合并到 trees[0] 中。\n删除 trees[1] ，trees = [[3,2,5,1,null,4]] 。\n\"\"\n结果树如上图所示，为一棵有效的二叉搜索树，所以返回该树的根节点。
\n\n

示例 2：

\n\"\"\n
\n输入：trees = [[5,3,8],[3,2,6]]\n输出：[]\n解释：\n选出 i=0 和 j=1 ，然后将 trees[1] 合并到 trees[0] 中。\n删除 trees[1] ，trees = [[5,3,8,2,6]] 。\n\"\"\n结果树如上图所示。仅能执行一次有效的操作，但结果树不是一棵有效的二叉搜索树，所以返回 null 。\n
\n\n

示例 3：

\n\"\"\n
\n输入：trees = [[5,4],[3]]\n输出：[]\n解释：无法执行任何操作。\n
\n\n

 

\n\n

提示：

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0 : val)\n * this.left = (left===undefined ? null : left)\n * this.right = (right===undefined ? null : right)\n * }\n */\n/**\n * @param {TreeNode[]} trees\n * @return {TreeNode}\n */\nvar canMerge = function(trees) {\n \n};", "__typename": "CodeSnippetNode" }, { "lang": "TypeScript", "langSlug": "typescript", "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\nfunction canMerge(trees: Array): TreeNode | null {\n\n};", "__typename": "CodeSnippetNode" }, { "lang": "PHP", "langSlug": "php", "code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * public $val = null;\n * public $left = null;\n * public $right = null;\n * function __construct($val = 0, $left = null, $right = null) {\n * $this->val = $val;\n * $this->left = $left;\n * $this->right = $right;\n * }\n * }\n */\nclass Solution {\n\n /**\n * @param TreeNode[] $trees\n * @return TreeNode\n */\n function canMerge($trees) {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Swift", "langSlug": "swift", "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 canMerge(_ trees: [TreeNode?]) -> TreeNode? {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "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 canMerge(trees: List): TreeNode? {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Go", "langSlug": "golang", "code": "/**\n * Definition for a binary tree node.\n * type TreeNode struct {\n * Val int\n * Left *TreeNode\n * Right *TreeNode\n * }\n */\nfunc canMerge(trees []*TreeNode) *TreeNode {\n \n}", "__typename": "CodeSnippetNode" }, { "lang": "Ruby", "langSlug": "ruby", "code": "# Definition for a binary tree node.\n# class TreeNode\n# attr_accessor :val, :left, :right\n# def initialize(val = 0, left = nil, right = nil)\n# @val = val\n# @left = left\n# @right = right\n# end\n# end\n# @param {TreeNode[]} trees\n# @return {TreeNode}\ndef can_merge(trees)\n \nend", "__typename": "CodeSnippetNode" }, { "lang": "Scala", "langSlug": "scala", "code": "/**\n * Definition for a binary tree node.\n * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {\n * var value: Int = _value\n * var left: TreeNode = _left\n * var right: TreeNode = _right\n * }\n */\nobject Solution {\n def canMerge(trees: List[TreeNode]): TreeNode = {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Rust", "langSlug": "rust", "code": "// Definition for a binary tree node.\n// #[derive(Debug, PartialEq, Eq)]\n// pub struct TreeNode {\n// pub val: i32,\n// pub left: Option>>,\n// pub right: Option>>,\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 can_merge(trees: Vec>>>) -> Option>> {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Racket", "langSlug": "racket", "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 (can-merge trees)\n (-> (listof (or/c tree-node? #f)) (or/c tree-node? #f))\n\n )", "__typename": "CodeSnippetNode" } ], "stats": "{\"totalAccepted\": \"2.3K\", \"totalSubmission\": \"6.7K\", \"totalAcceptedRaw\": 2256, \"totalSubmissionRaw\": 6722, \"acRate\": \"33.6%\"}", "hints": [ "Is it possible to have multiple leaf nodes with the same values?", "How many possible positions are there for each tree?", "The root value of the final tree does not occur as a value in any of the leaves of the original tree." ], "solution": null, "status": null, "sampleTestCase": "[[2,1],[3,2,5],[5,4]]", "metaData": "{\n \"name\": \"canMerge\",\n \"params\": [\n {\n \"name\": \"trees\",\n \"type\": \"list\"\n }\n ],\n \"return\": {\n \"type\": \"TreeNode\"\n },\n \"manual\": true\n}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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