mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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210 lines
26 KiB
JSON
210 lines
26 KiB
JSON
{
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"data": {
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"question": {
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"questionId": "1000019",
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"questionFrontendId": "面试题 17.12",
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"categoryTitle": "LCCI",
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"boundTopicId": 92975,
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"title": "BiNode LCCI",
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"titleSlug": "binode-lcci",
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"content": "<p>The data structure <code>TreeNode</code> 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 <code>TreeNode</code>) 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).</p>\r\n\r\n<p>Return the head node of the linked list after converting.</p>\r\n\r\n<p><b>Note: </b>This problem is slightly different from the original one in the book.</p>\r\n\r\n<p> </p>\r\n\r\n<p><strong>Example: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong> [4,2,5,1,3,null,6,0]\r\n<strong>Output: </strong> [0,null,1,null,2,null,3,null,4,null,5,null,6]\r\n</pre>\r\n\r\n<p><strong>Note: </strong></p>\r\n\r\n<ul>\r\n\t<li>The number of nodes will not exceed 100000.</li>\r\n</ul>\r\n",
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"translatedTitle": "BiNode",
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"translatedContent": "<p>二叉树数据结构<code>TreeNode</code>可用来表示单向链表（其中<code>left</code>置空，<code>right</code>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。</p>\n\n<p>返回转换后的单向链表的头节点。</p>\n\n<p><strong>注意：</strong>本题相对原题稍作改动</p>\n\n<p> </p>\n\n<p><strong>示例：</strong></p>\n\n<pre><strong>输入：</strong> [4,2,5,1,3,null,6,0]\n<strong>输出：</strong> [0,null,1,null,2,null,3,null,4,null,5,null,6]\n</pre>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>节点数量不会超过 100000。</li>\n</ul>\n",
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"isPaidOnly": false,
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"difficulty": "Easy",
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"likes": 104,
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"dislikes": 0,
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"isLiked": null,
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"similarQuestions": "[]",
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"contributors": [],
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"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}",
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"topicTags": [
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{
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"name": "Stack",
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"slug": "stack",
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"translatedName": "栈",
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"__typename": "TopicTagNode"
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},
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{
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"name": "Tree",
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"slug": "tree",
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"translatedName": "树",
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"__typename": "TopicTagNode"
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},
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{
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"name": "Depth-First Search",
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"slug": "depth-first-search",
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"translatedName": "深度优先搜索",
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"__typename": "TopicTagNode"
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},
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{
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"name": "Binary Search Tree",
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"slug": "binary-search-tree",
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"translatedName": "二叉搜索树",
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"__typename": "TopicTagNode"
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},
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{
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"name": "Linked List",
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"slug": "linked-list",
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"translatedName": "链表",
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"__typename": "TopicTagNode"
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},
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{
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"name": "Binary Tree",
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"slug": "binary-tree",
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"translatedName": "二叉树",
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"__typename": "TopicTagNode"
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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(int x) : val(x), left(NULL), right(NULL) {}\n * };\n */\nclass Solution {\npublic:\n TreeNode* convertBiNode(TreeNode* root) {\n\n }\n};",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Java",
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"langSlug": "java",
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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 convertBiNode(TreeNode root) {\n\n }\n}",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Python",
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"langSlug": "python",
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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 convertBiNode(self, root):\n \"\"\"\n :type root: TreeNode\n :rtype: TreeNode\n \"\"\"",
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"__typename": "CodeSnippetNode"
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},
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"lang": "Python3",
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"langSlug": "python3",
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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 convertBiNode(self, root: TreeNode) -> TreeNode:",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "C",
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"langSlug": "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 */\n\n\nstruct TreeNode* convertBiNode(struct TreeNode* root){\n\n}\n",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "C#",
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"langSlug": "csharp",
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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 x) { val = x; }\n * }\n */\npublic class Solution {\n public TreeNode ConvertBiNode(TreeNode root) {\n\n }\n}",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "JavaScript",
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"langSlug": "javascript",
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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 convertBiNode = function(root) {\n\n};",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Ruby",
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"langSlug": "ruby",
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"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",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Swift",
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"langSlug": "swift",
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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(_ 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}",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Go",
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"langSlug": "golang",
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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 convertBiNode(root *TreeNode) *TreeNode {\n\n}",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Scala",
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"langSlug": "scala",
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"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}",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Kotlin",
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"langSlug": "kotlin",
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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 convertBiNode(root: TreeNode?): TreeNode? {\n\n }\n}",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Rust",
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"langSlug": "rust",
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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 convert_bi_node(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {\n\n }\n}",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "PHP",
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"langSlug": "php",
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"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}",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "TypeScript",
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"langSlug": "typescript",
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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\nfunction convertBiNode(root: TreeNode | null): TreeNode | null {\n\n};",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Racket",
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"langSlug": "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 (convert-bi-node root)\n (-> (or/c tree-node? #f) (or/c tree-node? #f))\n\n )",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Erlang",
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"langSlug": "erlang",
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"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 .",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "Elixir",
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"langSlug": "elixir",
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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 convert_bi_node(root :: TreeNode.t | nil) :: TreeNode.t | nil\n def convert_bi_node(root) do\n\n end\nend",
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"__typename": "CodeSnippetNode"
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"hints": [
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"尝试递归解法。",
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"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？",
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"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。",
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"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。",
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"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？",
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"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？"
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],
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"solution": null,
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"sampleTestCase": "[4,2,5,1,3,null,6,0]",
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