{ "data": { "question": { "questionId": "1000019", "questionFrontendId": "面试题 17.12", "categoryTitle": "LCCI", "boundTopicId": 92975, "title": "BiNode LCCI", "titleSlug": "binode-lcci", "content": "

The data structure TreeNode is used for binary tree, but it can also used to represent a single linked list (where left is null, and right is the next node in the list). Implement a method to convert a binary search tree (implemented with TreeNode) into a single linked list. The values should be kept in order and the operation should be performed in place (that is, on the original data structure).

\r\n\r\n

Return the head node of the linked list after converting.

\r\n\r\n

Note: This problem is slightly different from the original one in the book.

\r\n\r\n

 

\r\n\r\n

Example:

\r\n\r\n
\r\nInput:  [4,2,5,1,3,null,6,0]\r\nOutput:  [0,null,1,null,2,null,3,null,4,null,5,null,6]\r\n
\r\n\r\n

Note:

\r\n\r\n\r\n", "translatedTitle": "BiNode", "translatedContent": "

二叉树数据结构TreeNode可用来表示单向链表（其中left置空，right为下一个链表节点）。实现一个方法，把二叉搜索树转换为单向链表，要求依然符合二叉搜索树的性质，转换操作应是原址的，也就是在原始的二叉搜索树上直接修改。

\n\n

返回转换后的单向链表的头节点。

\n\n

注意：本题相对原题稍作改动

\n\n

 

\n\n

示例：

\n\n
输入： [4,2,5,1,3,null,6,0]\n输出： [0,null,1,null,2,null,3,null,4,null,5,null,6]\n
\n\n

提示：

\n\n\n", "isPaidOnly": false, "difficulty": "Easy", "likes": 104, "dislikes": 0, "isLiked": null, "similarQuestions": "[]", "contributors": [], "langToValidPlayground": "{\"cpp\": true, \"java\": true, \"python\": true, \"python3\": true, \"mysql\": false, \"mssql\": false, \"oraclesql\": false, \"c\": false, \"csharp\": false, \"javascript\": false, \"ruby\": false, \"bash\": false, \"swift\": false, \"golang\": false, \"scala\": false, \"html\": false, \"pythonml\": false, \"kotlin\": false, \"rust\": false, \"php\": false, \"typescript\": false, \"racket\": false, \"erlang\": false, \"elixir\": false}", "topicTags": [ { "name": "Stack", "slug": "stack", "translatedName": "栈", "__typename": "TopicTagNode" }, { "name": "Tree", "slug": "tree", "translatedName": "树", "__typename": "TopicTagNode" }, { "name": "Depth-First Search", "slug": "depth-first-search", "translatedName": "深度优先搜索", "__typename": "TopicTagNode" }, { "name": "Binary Search Tree", "slug": "binary-search-tree", "translatedName": "二叉搜索树", "__typename": "TopicTagNode" }, { "name": "Linked List", "slug": "linked-list", "translatedName": "链表", "__typename": "TopicTagNode" }, { "name": "Binary Tree", "slug": "binary-tree", "translatedName": "二叉树", "__typename": "TopicTagNode" } ], "companyTagStats": null, "codeSnippets": [ { "lang": "C++", "langSlug": "cpp", "code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * TreeNode *left;\n * TreeNode *right;\n * TreeNode(int x) : val(x), left(NULL), right(NULL) {}\n * };\n */\nclass Solution {\npublic:\n TreeNode* convertBiNode(TreeNode* root) {\n\n }\n};", "__typename": "CodeSnippetNode" }, { "lang": "Java", "langSlug": "java", "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 convertBiNode(TreeNode root) {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Python", "langSlug": "python", "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 convertBiNode(self, root):\n \"\"\"\n :type root: TreeNode\n :rtype: TreeNode\n \"\"\"", "__typename": "CodeSnippetNode" }, { "lang": "Python3", "langSlug": "python3", "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 convertBiNode(self, root: TreeNode) -> TreeNode:", "__typename": "CodeSnippetNode" }, { "lang": "C", "langSlug": "c", "code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * struct TreeNode *left;\n * struct TreeNode *right;\n * };\n */\n\n\nstruct TreeNode* convertBiNode(struct TreeNode* root){\n\n}\n", "__typename": "CodeSnippetNode" }, { "lang": "C#", "langSlug": "csharp", "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 x) { val = x; }\n * }\n */\npublic class Solution {\n public TreeNode ConvertBiNode(TreeNode root) {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "JavaScript", "langSlug": "javascript", "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 convertBiNode = function(root) {\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)\n# @val = val\n# @left, @right = nil, nil\n# end\n# end\n\n# @param {TreeNode} root\n# @return {TreeNode}\ndef convert_bi_node(root)\n\nend", "__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(_ val: Int) {\n * self.val = val\n * self.left = nil\n * self.right = nil\n * }\n * }\n */\nclass Solution {\n func convertBiNode(_ root: TreeNode?) -> 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 convertBiNode(root *TreeNode) *TreeNode {\n\n}", "__typename": "CodeSnippetNode" }, { "lang": "Scala", "langSlug": "scala", "code": "/**\n * Definition for a binary tree node.\n * class TreeNode(var _value: Int) {\n * var value: Int = _value\n * var left: TreeNode = null\n * var right: TreeNode = null\n * }\n */\nobject Solution {\n def convertBiNode(root: 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 convertBiNode(root: 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 convert_bi_node(root: Option>>) -> Option>> {\n\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 $right = null;\n * function __construct($value) { $this->val = $value; }\n * }\n */\nclass Solution {\n\n /**\n * @param TreeNode $root\n * @return TreeNode\n */\n function convertBiNode($root) {\n\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 convertBiNode(root: TreeNode | null): TreeNode | null {\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 (convert-bi-node root)\n (-> (or/c tree-node? #f) (or/c tree-node? #f))\n\n )", "__typename": "CodeSnippetNode" }, { "lang": "Erlang", "langSlug": "erlang", "code": "%% Definition for a binary tree node.\n%%\n%% -record(tree_node, {val = 0 :: integer(),\n%% left = null :: 'null' | #tree_node{},\n%% right = null :: 'null' | #tree_node{}}).\n\n-spec convert_bi_node(Root :: #tree_node{} | null) -> #tree_node{} | null.\nconvert_bi_node(Root) ->\n .", "__typename": "CodeSnippetNode" }, { "lang": "Elixir", "langSlug": "elixir", "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 convert_bi_node(root :: TreeNode.t | nil) :: TreeNode.t | nil\n def convert_bi_node(root) do\n\n end\nend", "__typename": "CodeSnippetNode" } ], "stats": "{\"totalAccepted\": \"25.1K\", \"totalSubmission\": \"39.5K\", \"totalAcceptedRaw\": 25081, \"totalSubmissionRaw\": 39499, \"acRate\": \"63.5%\"}", "hints": [ "尝试递归解法。", "这样想：如果你有convertLeft和convertRight方法（它们可以把左右子树转换成双链表），你能使用它们把整个树转换成双链表吗？", "一旦你对递归算法有了一个基本的概念，就可能会陷入这种情况：有时你的递归算法需要返回链表的头部，有时它需要返回链表的尾部。解决这个问题有多种方法，想想不同的方法。", "要处理递归算法是返回链表的头节点还是尾节点，可以尝试传递一个参数作为标志。但这不会很好。问题是，当调用convert(current.left)时，你希望得到left链表的尾节点。这样就可以将链表的末尾与current连接。但是，如果current是其他节点的右子树，那么convert(current)需要返回链表的头节点（其实是current.left的头节点）。实际上，链表的头节点和尾节点你都需要。", "许多人在这一点上左右为难，不知道该怎么办。有时他们需要链表的头部，有时他们需要链表的尾部。给定的节点通常不知道它在convert调用中应返回什么。有时候，最简单的解决方案就是：总是同时返回这两个值。有什么方法可以做到这一点？", "可以通过多种方式返回链表的头部和尾部。可以返回一个双元素数组，可以定义一个新的数据结构来保存头节点和尾节点，还可以重用BiNode数据结构。如果你使用的语言（如Python）支持返回多个值，你就可以使用此功能。可以将这个问题作为一个循环链表来解决，即头节点的前一个指针指向尾部，然后在外部的函数中拆开循环链表。试试这些解决方案。你最喜欢哪个？为什么？" ], "solution": null, "status": null, "sampleTestCase": "[4,2,5,1,3,null,6,0]", "metaData": "{\"name\": \"convertBiNode\", \"params\": [{\"name\": \"root\", \"type\": \"TreeNode\"}], \"return\": {\"type\": \"TreeNode\"}}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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