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{
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"question": {
"questionId": "1000568",
"questionFrontendId": "LCP 80",
"categoryTitle": "Algorithms",
"boundTopicId": 2258736,
"title": "生物进化录",
"titleSlug": "qoQAMX",
"content": "English description is not available for the problem. Please switch to Chinese.",
"translatedTitle": "生物进化录",
"translatedContent": "在永恒之森中,存在着一本生物进化录,以 **一个树形结构** 记载了所有生物的演化过程。经过观察并整理了各节点间的关系,`parents[i]` 表示编号 `i` 节点的父节点编号(根节点的父节点为 `-1`)。\n\n为了探索和记录其中的演化规律队伍中的炼金术师提出了一种方法可以以字符串的形式将其复刻下来规则如下\n- 初始只有一个根节点,表示演化的起点,依次记录 `01` 字符串中的字符,\n- 如果记录 `0`,则在当前节点下添加一个子节点,并将指针指向新添加的子节点;\n- 如果记录 `1`,则将指针回退到当前节点的父节点处。\n\n现在需要应用上述的记录方法复刻下它的演化过程。请返回能够复刻演化过程的字符串中 **字典序最小** 的 `01` 字符串。\n\n**注意:**\n- 节点指针最终可以停在任何节点上,不一定要回到根节点。\n\n**示例 1**\n> 输入:`parents = [-1,0,0,2]`\n>\n> 输出:`\"00110\"`\n>\n>解释:树结构如下图所示,共存在 2 种记录方案:\n>第 1 种方案为0(记录编号 1 的节点) -> 1(回退至节点 0) -> 0(记录编号 2 的节点) -> 0((记录编号 3 的节点))\n>第 2 种方案为0(记录编号 2 的节点) -> 0(记录编号 3 的节点) -> 1(回退至节点 2) -> 1(回退至节点 0) -> 0(记录编号 1 的节点)\n>返回字典序更小的 `\"00110\"`\n![image.png](https://pic.leetcode.cn/1682319485-cRVudI-image.png){:width=120px}![进化 (3).gif](https://pic.leetcode.cn/1682412701-waHdnm-%E8%BF%9B%E5%8C%96%20\\(3\\).gif){:width=320px}\n\n\n\n**示例 2**\n> 输入:`parents = [-1,0,0,1,2,2]`\n>\n> 输出:`\"00101100\"`\n\n**提示:**\n\n- `1 <= parents.length <= 10^4`\n- `-1 <= parents[i] < i` (即父节点编号小于子节点)",
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"difficulty": "Hard",
"likes": 4,
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{
"lang": "C++",
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"code": "class Solution {\npublic:\n string evolutionaryRecord(vector<int>& parents) {\n\n }\n};",
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"lang": "Java",
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"code": "class Solution {\n public String evolutionaryRecord(int[] parents) {\n\n }\n}",
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{
"lang": "Python",
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"code": "class Solution(object):\n def evolutionaryRecord(self, parents):\n \"\"\"\n :type parents: List[int]\n :rtype: str\n \"\"\"",
"__typename": "CodeSnippetNode"
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{
"lang": "Python3",
"langSlug": "python3",
"code": "class Solution:\n def evolutionaryRecord(self, parents: List[int]) -> str:",
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"code": "char* evolutionaryRecord(int* parents, int parentsSize){\n\n}",
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"code": "/**\n * @param {number[]} parents\n * @return {string}\n */\nvar evolutionaryRecord = function(parents) {\n\n};",
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