{ "data": { "question": { "questionId": "100229", "questionFrontendId": "面试题 04.10", "categoryTitle": "LCCI", "boundTopicId": 58158, "title": "Check SubTree LCCI", "titleSlug": "check-subtree-lcci", "content": "

T1 and T2 are two very large binary trees. Create an algorithm to determine if T2 is a subtree of T1.

\r\n\r\n

A tree T2 is a subtree of T1 if there exists a node n in T1 such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical.

\r\n\r\n

Note: This problem is slightly different from the original problem.

\r\n\r\n

Example1:

\r\n\r\n
\r\n Input: t1 = [1, 2, 3], t2 = [2]\r\n Output: true\r\n
\r\n\r\n

Example2:

\r\n\r\n
\r\n Input: t1 = [1, null, 2, 4], t2 = [3, 2]\r\n Output: false\r\n
\r\n\r\n

Note:

\r\n\r\n
    \r\n\t
  1. The node numbers of both tree are in [0, 20000].
  2. \r\n
\r\n", "translatedTitle": "检查子树", "translatedContent": "

检查子树。你有两棵非常大的二叉树:T1,有几万个节点;T2,有几万个节点。设计一个算法,判断 T2 是否为 T1 的子树。

\n\n

如果 T1 有这么一个节点 n,其子树与 T2 一模一样,则 T2 为 T1 的子树,也就是说,从节点 n 处把树砍断,得到的树与 T2 完全相同。

\n\n

注意:此题相对书上原题略有改动。

\n\n

示例1:

\n\n
\n 输入:t1 = [1, 2, 3], t2 = [2]\n 输出:true\n
\n\n

示例2:

\n\n
\n 输入:t1 = [1, null, 2, 4], t2 = [3, 2]\n 输出:false\n
\n\n

提示:

\n\n
    \n\t
  1. 树的节点数目范围为[0, 20000]。
  2. \n
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0 : val)\n * this.left = (left===undefined ? null : left)\n * this.right = (right===undefined ? null : right)\n * }\n * }\n */\n\nfunction checkSubTree(t1: TreeNode | null, t2: TreeNode | null): boolean {\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 $t1\n * @param TreeNode $t2\n * @return Boolean\n */\n function checkSubTree($t1, $t2) {\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(_ val: Int) {\n * self.val = val\n * self.left = nil\n * self.right = nil\n * }\n * }\n */\nclass Solution {\n func checkSubTree(_ t1: TreeNode?, _ t2: TreeNode?) -> Bool {\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 checkSubTree(t1: TreeNode?, t2: TreeNode?): Boolean {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Dart", "langSlug": "dart", "code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * int val;\n * TreeNode? left;\n * TreeNode? right;\n * TreeNode([this.val = 0, this.left, this.right]);\n * }\n */\nclass Solution {\n bool checkSubTree(TreeNode? t1, TreeNode? t2) {\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 checkSubTree(t1 *TreeNode, t2 *TreeNode) bool {\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} t1\n# @param {TreeNode} t2\n# @return {Boolean}\ndef check_sub_tree(t1, t2)\n\nend", "__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 checkSubTree(t1: TreeNode, t2: TreeNode): Boolean = {\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 check_sub_tree(t1: Option>>, t2: Option>>) -> bool {\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 (check-sub-tree t1 t2)\n (-> (or/c tree-node? #f) (or/c tree-node? #f) boolean?)\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 check_sub_tree(T1 :: #tree_node{} | null, T2 :: #tree_node{} | null) -> boolean().\ncheck_sub_tree(T1, T2) ->\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 check_sub_tree(t1 :: TreeNode.t | nil, t2 :: TreeNode.t | nil) :: boolean\n def check_sub_tree(t1, t2) do\n\n end\nend", "__typename": "CodeSnippetNode" } ], "stats": "{\"totalAccepted\": \"28.4K\", \"totalSubmission\": \"43.1K\", \"totalAcceptedRaw\": 28433, \"totalSubmissionRaw\": 43070, \"acRate\": \"66.0%\"}", "hints": [ "如果T2是T1的子树,它的中序遍历将如何与T1的比较?它的前序和后序遍历如何?", "中序遍历无法告诉我们更多。毕竟,每个具有相同值的二叉搜索树(不管结构如何)将具有相同的中序遍历。这也就是中序遍历的含义:内容是有序的(如果它在二叉搜索树这种特定情况下不起作用,那么对于一般二叉树来说它肯定不起作用)。然而,前序遍历更具指示性。", "你可能得出结论,如果T2.preorderTraversal()是T1.preorderTraversal()的子字符串,则T2是T1的子树。这几乎是事实,除非树可能有重复的值。假设T1和T2具有所有重复值,但结构不同。即使T2不是T1的子树,前序遍历看起来也是一样的。你如何处理这样的情况?", "尽管问题似乎源于重复的值,但不止如此。问题是,前序遍历是相同的,只是因为我们跳过了空节点(因为它们是空的)。考虑在访问到空节点时往前序遍历的字符串中插入一个占位符。把空节点记录为一个“真正的”节点,你就可以区分出不同的结构了。", "或者用递归法处理这个问题。给定一个特殊节点T1,可以检查它的子树是否匹配T2吗?" ], "solution": null, "status": null, "sampleTestCase": "[1, 2, 3]\n[2]", "metaData": "{\"name\": \"checkSubTree\", \"params\": [{\"name\": \"t1\", \"type\": \"TreeNode\"}, {\"name\": \"t2\", \"type\": \"TreeNode\"}], \"return\": {\"type\": \"boolean\"}}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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