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leetcode-problemset/leetcode-cn/originData/shu-de-zi-jie-gou-lcof.json
2022-05-02 23:44:12 +08:00

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{
"data": {
"question": {
"questionId": "100287",
"questionFrontendId": "剑指 Offer 26",
"categoryTitle": "LCOF",
"boundTopicId": 85601,
"title": "树的子结构 LCOF",
"titleSlug": "shu-de-zi-jie-gou-lcof",
"content": "English description is not available for the problem. Please switch to Chinese.",
"translatedTitle": "树的子结构",
"translatedContent": "<p>输入两棵二叉树A和B判断B是不是A的子结构。(约定空树不是任意一个树的子结构)</p>\n\n<p>B是A的子结构 即 A中有出现和B相同的结构和节点值。</p>\n\n<p>例如:<br>\n给定的树 A:</p>\n\n<p><code>&nbsp; &nbsp; &nbsp;3<br>\n&nbsp; &nbsp; / \\<br>\n&nbsp; &nbsp;4 &nbsp; 5<br>\n&nbsp; / \\<br>\n&nbsp;1 &nbsp; 2</code><br>\n给定的树 B</p>\n\n<p><code>&nbsp; &nbsp;4&nbsp;<br>\n&nbsp; /<br>\n&nbsp;1</code><br>\n返回 true因为 B 与 A 的一个子树拥有相同的结构和节点值。</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre><strong>输入:</strong>A = [1,2,3], B = [3,1]\n<strong>输出:</strong>false\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre><strong>输入:</strong>A = [3,4,5,1,2], B = [4,1]\n<strong>输出:</strong>true</pre>\n\n<p><strong>限制:</strong></p>\n\n<p><code>0 &lt;= 节点个数 &lt;= 10000</code></p>\n",
"isPaidOnly": false,
"difficulty": "Medium",
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{
"name": "Tree",
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"translatedName": "树",
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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 boolean isSubStructure(TreeNode A, TreeNode B) {\n\n }\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 isSubStructure(self, A: TreeNode, B: 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 isSubStructure(struct TreeNode* A, struct TreeNode* B){\n\n}\n",
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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 bool IsSubStructure(TreeNode A, TreeNode B) {\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} A\n * @param {TreeNode} B\n * @return {boolean}\n */\nvar isSubStructure = function(A, B) {\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 isSubStructure(A: TreeNode?, B: 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 is_sub_structure(a: Option<Rc<RefCell<TreeNode>>>, b: 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 (is-sub-structure A B)\n (-> (or/c tree-node? #f) (or/c tree-node? #f) boolean?)\n\n )",
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