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leetcode-problemset/leetcode-cn/originData/convert-sorted-list-to-binary-search-tree.json
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{
"data": {
"question": {
"questionId": "109",
"questionFrontendId": "109",
"categoryTitle": "Algorithms",
"boundTopicId": 1294,
"title": "Convert Sorted List to Binary Search Tree",
"titleSlug": "convert-sorted-list-to-binary-search-tree",
"content": "<p>Given the <code>head</code> of a singly linked list where elements are sorted in <strong>ascending order</strong>, convert <em>it to a </em><span data-keyword=\"height-balanced\"><strong><em>height-balanced</em></strong></span> <em>binary search tree</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/2020/08/17/linked.jpg\" style=\"width: 500px; height: 388px;\" />\n<pre>\n<strong>Input:</strong> head = [-10,-3,0,5,9]\n<strong>Output:</strong> [0,-3,9,-10,null,5]\n<strong>Explanation:</strong> One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> head = []\n<strong>Output:</strong> []\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in <code>head</code> is in the range <code>[0, 2 * 10<sup>4</sup>]</code>.</li>\n\t<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
"translatedTitle": "有序链表转换二叉搜索树",
"translatedContent": "<p>给定一个单链表的头节点 &nbsp;<code>head</code>&nbsp;,其中的元素 <strong>按升序排序</strong> ,将其转换为高度平衡的二叉搜索树。</p>\n\n<p>本题中,一个高度平衡二叉树是指一个二叉树<em>每个节点&nbsp;</em>的左右两个子树的高度差不超过 1。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1:</strong></p>\n\n<p><img src=\"https://assets.leetcode.com/uploads/2020/08/17/linked.jpg\" style=\"height: 388px; width: 500px;\" /></p>\n\n<pre>\n<strong>输入:</strong> head = [-10,-3,0,5,9]\n<strong>输出:</strong> [0,-3,9,-10,null,5]\n<strong>解释:</strong> 一个可能的答案是[0-3,9-10,null,5],它表示所示的高度平衡的二叉搜索树。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong> head = []\n<strong>输出:</strong> []\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>head</code>&nbsp;中的节点数在<code>[0, 2 * 10<sup>4</sup>]</code>&nbsp;范围内</li>\n\t<li><code>-10<sup>5</sup>&nbsp;&lt;= Node.val &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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"code": "/**\n * Definition for singly-linked list.\n * type ListNode struct {\n * Val int\n * Next *ListNode\n * }\n */\n/**\n * Definition for a binary tree node.\n * type TreeNode struct {\n * Val int\n * Left *TreeNode\n * Right *TreeNode\n * }\n */\nfunc sortedListToBST(head *ListNode) *TreeNode {\n\n}",
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"code": "# Definition for singly-linked list.\n# class ListNode\n# attr_accessor :val, :next\n# def initialize(val = 0, _next = nil)\n# @val = val\n# @next = _next\n# end\n# end\n# 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 {ListNode} head\n# @return {TreeNode}\ndef sorted_list_to_bst(head)\n\nend",
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"code": "/**\n * Definition for singly-linked list.\n * class ListNode(_x: Int = 0, _next: ListNode = null) {\n * var next: ListNode = _next\n * var x: Int = _x\n * }\n */\n/**\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 sortedListToBST(head: ListNode): TreeNode = {\n\n }\n}",
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"code": "// Definition for singly-linked list.\n// #[derive(PartialEq, Eq, Clone, Debug)]\n// pub struct ListNode {\n// pub val: i32,\n// pub next: Option<Box<ListNode>>\n// }\n//\n// impl ListNode {\n// #[inline]\n// fn new(val: i32) -> Self {\n// ListNode {\n// next: None,\n// val\n// }\n// }\n// }\n// 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 sorted_list_to_bst(head: Option<Box<ListNode>>) -> Option<Rc<RefCell<TreeNode>>> {\n\n }\n}",
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"code": "; Definition for singly-linked list:\n#|\n\n; val : integer?\n; next : (or/c list-node? #f)\n(struct list-node\n (val next) #:mutable #:transparent)\n\n; constructor\n(define (make-list-node [val 0])\n (list-node val #f))\n\n|#\n\n; 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 (sorted-list-to-bst head)\n (-> (or/c list-node? #f) (or/c tree-node? #f))\n )",
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"code": "%% Definition for singly-linked list.\n%%\n%% -record(list_node, {val = 0 :: integer(),\n%% next = null :: 'null' | #list_node{}}).\n\n%% 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 sorted_list_to_bst(Head :: #list_node{} | null) -> #tree_node{} | null.\nsorted_list_to_bst(Head) ->\n .",
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"code": "# Definition for singly-linked list.\n#\n# defmodule ListNode do\n# @type t :: %__MODULE__{\n# val: integer,\n# next: ListNode.t() | nil\n# }\n# defstruct val: 0, next: nil\n# end\n\n# 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 sorted_list_to_bst(head :: ListNode.t | nil) :: TreeNode.t | nil\n def sorted_list_to_bst(head) do\n \n end\nend",
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