leetcode-problemset/leetcode-cn/originData/check-subtree-lcci.json

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            "questionId": "100229",
            "questionFrontendId": "面试题 04.10",
            "categoryTitle": "LCCI",
            "boundTopicId": 58158,
            "title": "Check SubTree LCCI",
            "titleSlug": "check-subtree-lcci",
            "content": "<p>T1&nbsp;and T2 are two very large binary trees. Create an algorithm to determine if T2 is a subtree of T1.</p>\r\n\r\n<p>A tree T2 is a subtree of T1&nbsp;if there exists a node n in T1&nbsp;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.</p>\r\n\r\n<p><strong>Note:</strong> This problem is slightly different from the original problem.</p>\r\n\r\n<p><strong>Example1:</strong></p>\r\n\r\n<pre>\r\n<strong> Input</strong>: t1 = [1, 2, 3], t2 = [2]\r\n<strong> Output</strong>: true\r\n</pre>\r\n\r\n<p><strong>Example2:</strong></p>\r\n\r\n<pre>\r\n<strong> Input</strong>: t1 = [1, null, 2, 4], t2 = [3, 2]\r\n<strong> Output</strong>: false\r\n</pre>\r\n\r\n<p><strong>Note: </strong></p>\r\n\r\n<ol>\r\n\t<li>The node numbers of both tree are in [0, 20000].</li>\r\n</ol>\r\n",
            "translatedTitle": "检查子树",
            "translatedContent": "<p>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。</p>\n\n<p>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。</p>\n\n<p><strong>注意：</strong>此题相对书上原题略有改动。</p>\n\n<p><strong>示例1:</strong></p>\n\n<pre>\n<strong> 输入</strong>：t1 = [1, 2, 3], t2 = [2]\n<strong> 输出</strong>：true\n</pre>\n\n<p><strong>示例2:</strong></p>\n\n<pre>\n<strong> 输入</strong>：t1 = [1, null, 2, 4], t2 = [3, 2]\n<strong> 输出</strong>：false\n</pre>\n\n<p><strong>提示：</strong></p>\n\n<ol>\n\t<li>树的节点数目范围为[0, 20000]。</li>\n</ol>\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 checkSubTree(self, t1: TreeNode, t2: TreeNode) -> bool:",
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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\nbool checkSubTree(struct TreeNode* t1, struct TreeNode* t2){\n\n}\n",
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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 checkSubTree(t1: TreeNode | null, t2: TreeNode | null): boolean {\n\n};",
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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 checkSubTree(_ t1: TreeNode?, _ t2: TreeNode?) -> Bool {\n\n    }\n}",
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                    "lang": "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 checkSubTree(t1: TreeNode?, t2: TreeNode?): Boolean {\n\n    }\n}",
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                    "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}",
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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 checkSubTree(t1 *TreeNode, t2 *TreeNode) bool {\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} t1\n# @param {TreeNode} t2\n# @return {Boolean}\ndef check_sub_tree(t1, t2)\n\nend",
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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 checkSubTree(t1: TreeNode, t2: TreeNode): Boolean = {\n\n    }\n}",
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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 check_sub_tree(t1: Option<Rc<RefCell<TreeNode>>>, t2: Option<Rc<RefCell<TreeNode>>>) -> bool {\n\n    }\n}",
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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 (check-sub-tree t1 t2)\n  (-> (or/c tree-node? #f) (or/c tree-node? #f) boolean?)\n\n  )",
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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 check_sub_tree(T1 :: #tree_node{} | null, T2 :: #tree_node{} | null) -> boolean().\ncheck_sub_tree(T1, T2) ->\n  .",
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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 check_sub_tree(t1 :: TreeNode.t | nil, t2 :: TreeNode.t | nil) :: boolean\n  def check_sub_tree(t1, t2) do\n\n  end\nend",
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            "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？"
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