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"data": {
"question": {
"questionId": "1000304",
"questionFrontendId": "LCR 048",
"categoryTitle": "Algorithms",
"boundTopicId": 910298,
"title": "二叉树的序列化与反序列化",
"titleSlug": "h54YBf",
"content": "<p>English description is not available for the problem. Please switch to Chinese.</p>\n",
"translatedTitle": "二叉树的序列化与反序列化",
"translatedContent": "<p>序列化是将一个数据结构或者对象转换为连续的比特位的操作,进而可以将转换后的数据存储在一个文件或者内存中,同时也可以通过网络传输到另一个计算机环境,采取相反方式重构得到原数据。</p>\n\n<p>请设计一个算法来实现二叉树的序列化与反序列化。这里不限定你的序列 / 反序列化算法执行逻辑,只需要保证一个二叉树可以被序列化为一个字符串并且将这个字符串反序列化为原始的树结构。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/09/15/serdeser.jpg\" style=\"width: 442px; height: 324px;\" /></p>\n\n<pre>\n<strong>输入:</strong>root = [1,2,3,null,null,4,5]\n<strong>输出:</strong>[1,2,3,null,null,4,5]\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>root = []\n<strong>输出:</strong>[]\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<strong>输入:</strong>root = [1]\n<strong>输出:</strong>[1]\n</pre>\n\n<p><strong>示例 4</strong></p>\n\n<pre>\n<strong>输入:</strong>root = [1,2]\n<strong>输出:</strong>[1,2]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>输入输出格式与 LeetCode 目前使用的方式一致,详情请参阅&nbsp;<a href=\"/faq/#binary-tree\">LeetCode 序列化二叉树的格式</a>。你并非必须采取这种方式,也可以采用其他的方法解决这个问题。</li>\n\t<li>树中结点数在范围 <code>[0, 10<sup>4</sup>]</code> 内</li>\n\t<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><meta charset=\"UTF-8\" />注意:本题与主站 297&nbsp;题相同:<a href=\"https://leetcode-cn.com/problems/serialize-and-deserialize-binary-tree/\">https://leetcode-cn.com/problems/serialize-and-deserialize-binary-tree/</a>&nbsp;</p>\n",
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"code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * TreeNode *left;\n * TreeNode *right;\n * TreeNode(int x) : val(x), left(NULL), right(NULL) {}\n * };\n */\nclass Codec {\npublic:\n\n // Encodes a tree to a single string.\n string serialize(TreeNode* root) {\n \n }\n\n // Decodes your encoded data to tree.\n TreeNode* deserialize(string data) {\n \n }\n};\n\n// Your Codec object will be instantiated and called as such:\n// Codec ser, deser;\n// TreeNode* ans = deser.deserialize(ser.serialize(root));",
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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 */\npublic class Codec {\n\n // Encodes a tree to a single string.\n public String serialize(TreeNode root) {\n \n }\n\n // Decodes your encoded data to tree.\n public TreeNode deserialize(String data) {\n \n }\n}\n\n// Your Codec object will be instantiated and called as such:\n// Codec ser = new Codec();\n// Codec deser = new Codec();\n// TreeNode ans = deser.deserialize(ser.serialize(root));",
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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 Codec:\n\n def serialize(self, root):\n \"\"\"Encodes a tree to a single string.\n \n :type root: TreeNode\n :rtype: str\n \"\"\"\n \n\n def deserialize(self, data):\n \"\"\"Decodes your encoded data to tree.\n \n :type data: str\n :rtype: TreeNode\n \"\"\"\n \n\n# Your Codec object will be instantiated and called as such:\n# ser = Codec()\n# deser = Codec()\n# ans = deser.deserialize(ser.serialize(root))",
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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 Codec:\n\n def serialize(self, root):\n \"\"\"Encodes a tree to a single string.\n \n :type root: TreeNode\n :rtype: str\n \"\"\"\n \n\n def deserialize(self, data):\n \"\"\"Decodes your encoded data to tree.\n \n :type data: str\n :rtype: TreeNode\n \"\"\"\n \n\n# Your Codec object will be instantiated and called as such:\n# ser = Codec()\n# deser = Codec()\n# ans = deser.deserialize(ser.serialize(root))",
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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 Codec {\n\n // Encodes a tree to a single string.\n public string serialize(TreeNode root) {\n \n }\n\n // Decodes your encoded data to tree.\n public TreeNode deserialize(string data) {\n \n }\n}\n\n// Your Codec object will be instantiated and called as such:\n// Codec ser = new Codec();\n// Codec deser = new Codec();\n// TreeNode ans = deser.deserialize(ser.serialize(root));",
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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\n/*\n * Encodes a tree to a single string.\n */\nfunction serialize(root: TreeNode | null): string {\n\n};\n\n/*\n * Decodes your encoded data to tree.\n */\nfunction deserialize(data: string): TreeNode | null {\n\n};\n\n\n/**\n * Your functions will be called as such:\n * deserialize(serialize(root));\n */",
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"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 */\n\nclass Codec {\n function __construct() {\n \n }\n \n /**\n * @param TreeNode $root\n * @return String\n */\n function serialize($root) {\n \n }\n \n /**\n * @param String $data\n * @return TreeNode\n */\n function deserialize($data) {\n \n }\n}\n\n/**\n * Your Codec object will be instantiated and called as such:\n * $ser = Codec();\n * $deser = Codec();\n * $data = $ser->serialize($root);\n * $ans = $deser->deserialize($data);\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 */\n\nclass Codec {\n func serialize(_ root: TreeNode?) -> String {\n \n }\n \n func deserialize(_ data: String) -> TreeNode? {\n \n }\n}\n\n// Your Codec object will be instantiated and called as such:\n// var ser = Codec()\n// var deser = Codec()\n// deser.deserialize(ser.serialize(root))",
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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 */\n\ntype Codec struct {\n \n}\n\nfunc Constructor() Codec {\n \n}\n\n// Serializes a tree to a single string.\nfunc (this *Codec) serialize(root *TreeNode) string {\n \n}\n\n// Deserializes your encoded data to tree.\nfunc (this *Codec) deserialize(data string) *TreeNode { \n \n}\n\n\n/**\n * Your Codec object will be instantiated and called as such:\n * ser := Constructor();\n * deser := Constructor();\n * data := ser.serialize(root);\n * ans := deser.deserialize(data);\n */",
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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 */\n\nclass Codec {\n // Encodes a list of strings to a single string.\n def serialize(root: TreeNode): String = {\n \n }\n \n // Decodes a single string to a list of strings.\n def deserialize(data: String): TreeNode = {\n \n }\n}\n\n/**\n * Your Codec object will be instantiated and called as such:\n * var ser = new Codec()\n * var deser = new Codec()\n * val s = ser.serialize(root)\n * val ans = deser.deserialize(s)\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;\nstruct Codec {\n\t\n}\n\n/** \n * `&self` means the method takes an immutable reference.\n * If you need a mutable reference, change it to `&mut self` instead.\n */\nimpl Codec {\n fn new() -> Self {\n \n }\n\n fn serialize(&self, root: Option<Rc<RefCell<TreeNode>>>) -> String {\n \n }\n\t\n fn deserialize(&self, data: String) -> Option<Rc<RefCell<TreeNode>>> {\n \n }\n}\n\n/**\n * Your Codec object will be instantiated and called as such:\n * let obj = Codec::new();\n * let data: String = obj.serialize(strs);\n * let ans: Option<Rc<RefCell<TreeNode>>> = obj.deserialize(data);\n */",
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