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{
"data": {
"question": {
"questionId": "1000010",
"questionFrontendId": "面试题 04.09",
"categoryTitle": "LCCI",
"boundTopicId": 91535,
"title": "BST Sequences LCCI",
"titleSlug": "bst-sequences-lcci",
"content": "<p>A binary search tree was created by traversing through an array from left to right and inserting each element. Given a binary search tree with distinct elements, print all possible arrays that could have led to this tree.</p>\r\n\r\n<p><strong>Example:</strong><br />\r\nGiven the following tree:</p>\r\n\r\n<pre>\r\n 2\r\n / \\\r\n 1 3\r\n</pre>\r\n\r\n<p>Output:</p>\r\n\r\n<pre>\r\n[\r\n [2,1,3],\r\n [2,3,1]\r\n]\r\n</pre>\r\n",
"translatedTitle": "二叉搜索树序列",
"translatedContent": "<p>从左向右遍历一个数组,通过不断将其中的元素插入树中可以逐步地生成一棵二叉搜索树。</p>\n\n<p>给定一个由<strong>不同节点</strong>组成的二叉搜索树 <code>root</code>,输出所有可能生成此树的数组。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入: </strong>root = [2,1,3]\n<strong>输出: </strong>[[2,1,3],[2,3,1]]\n解释: 数组 [2,1,3]、[2,3,1] 均可以通过从左向右遍历元素插入树中形成以下二叉搜索树\n&nbsp; 2 \n&nbsp; / \\ \n&nbsp; 1 3\n</pre>\n\n<p><meta charset=\"UTF-8\" /></p>\n\n<p><strong>示例</strong><strong>&nbsp;2:</strong></p>\n\n<pre>\n<strong>输入: </strong>root = [4,1,null,null,3,2]\n<strong>输出: </strong>[[4,1,3,2]]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>二叉搜索树中的节点数在<meta charset=\"UTF-8\" />&nbsp;<code>[0, 1000]</code>&nbsp;的范围内</li>\n\t<li><code>1 &lt;= 节点值&nbsp;&lt;= 10^6</code></li>\n\t<li>\n\t<p>用例保证符合要求的数组数量不超过 <code>5000</code></p>\n\t</li>\n</ul>\n",
"isPaidOnly": false,
"difficulty": "Hard",
"likes": 89,
"dislikes": 0,
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{
"name": "Tree",
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"translatedName": "树",
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"translatedName": "二叉搜索树",
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{
"name": "Backtracking",
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"translatedName": "回溯",
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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 List<List<Integer>> BSTSequences(TreeNode root) {\n\n }\n}",
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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 BSTSequences(self, root):\n \"\"\"\n :type root: TreeNode\n :rtype: List[List[int]]\n \"\"\"",
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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 BSTSequences(self, root: TreeNode) -> List[List[int]]:",
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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\n/**\n * Return an array of arrays of size *returnSize.\n * The sizes of the arrays are returned as *returnColumnSizes array.\n * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().\n */\nint** BSTSequences(struct TreeNode* root, int* returnSize, int** returnColumnSizes){\n\n}\n",
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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 {number[][]}\n */\nvar BSTSequences = function(root) {\n\n};",
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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 {Integer[][]}\ndef bst_sequences(root)\n\nend",
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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 BSTSequences(_ root: TreeNode?) -> [[Int]] {\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 BSTSequences(root *TreeNode) [][]int {\n\n}",
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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 BSTSequences(root: TreeNode?): List<List<Int>> {\n\n }\n}",
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"lang": "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 bst_sequences(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {\n\n }\n}",
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"lang": "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 BSTSequences(root: TreeNode | null): number[][] {\n\n};",
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"lang": "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 (bst-sequences root)\n (-> (or/c tree-node? #f) (listof (listof exact-integer?)))\n\n )",
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"lang": "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 bst_sequences(Root :: #tree_node{} | null) -> [[integer()]].\nbst_sequences(Root) ->\n .",
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"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 bst_sequences(root :: TreeNode.t | nil) :: [[integer]]\n def bst_sequences(root) do\n\n end\nend",
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"hints": [
"每个数组中的第一个值是多少?",
"根是每个数组中必须包含的第一个值。相对于右子树中的值,左子树中的值顺序如何?左子树值是否需要在右子树之前插入?",
"左子树值与右子树值之间本质上可以是任何关系。可以在右子树之前插入左子树值,也可以反转(右子树的值在左边)或采用任意其他顺序。",
"把这个分解成子问题。使用递归法。如果你有左右子树的所有可能的序列,那么如何为整个树创建所有可能的序列呢?"
],
"solution": null,
"status": null,
"sampleTestCase": "[2,1,3]",
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