mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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140 lines
19 KiB
JSON
140 lines
19 KiB
JSON
{
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"data": {
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"question": {
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"questionId": "100179",
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"questionFrontendId": "面试题 04.08",
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"categoryTitle": "LCCI",
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"boundTopicId": 46238,
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"title": "First Common Ancestor LCCI",
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"titleSlug": "first-common-ancestor-lcci",
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"content": "<p>Design an algorithm and write code to find the first common ancestor of two nodes in a binary tree. Avoid storing additional nodes in a data structure. NOTE: This is not necessarily a binary search tree.</p>\r\n\r\n<p>For example, Given the following tree: root = [3,5,1,6,2,0,8,null,null,7,4]</p>\r\n\r\n<pre>\r\n 3\r\n / \\\r\n 5 1\r\n / \\ / \\\r\n6 2 0 8\r\n / \\\r\n 7 4\r\n</pre>\r\n\r\n<p><strong>Example 1:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1\r\n<strong>Input:</strong> 3\r\n<strong>Explanation:</strong> The first common ancestor of node 5 and node 1 is node 3.</pre>\r\n\r\n<p><strong>Example 2:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4\r\n<strong>Output:</strong> 5\r\n<strong>Explanation:</strong> The first common ancestor of node 5 and node 4 is node 5.</pre>\r\n\r\n<p><strong>Notes:</strong></p>\r\n\r\n<ul>\r\n\t<li>All node values are pairwise distinct.</li>\r\n\t<li>p, q are different node and both can be found in the given tree.</li>\r\n</ul>\r\n",
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"translatedTitle": "首个共同祖先",
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"translatedContent": "<p>设计并实现一个算法,找出二叉树中某两个节点的第一个共同祖先。不得将其他的节点存储在另外的数据结构中。注意:这不一定是二叉搜索树。</p>\n\n<p>例如,给定如下二叉树: root = [3,5,1,6,2,0,8,null,null,7,4]</p>\n\n<pre> 3\n / \\\n 5 1\n / \\ / \\\n6 2 0 8\n / \\\n 7 4\n</pre>\n\n<p><strong>示例 1:</strong></p>\n\n<pre><strong>输入:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1\n<strong>输出:</strong> 3\n<strong>解释:</strong> 节点 5 和节点 1 的最近公共祖先是节点 3。</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre><strong>输入:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4\n<strong>输出:</strong> 5\n<strong>解释:</strong> 节点 5 和节点 4 的最近公共祖先是节点 5。因为根据定义最近公共祖先节点可以为节点本身。</pre>\n\n<p><strong>说明:</strong></p>\n\n<pre>所有节点的值都是唯一的。\np、q 为不同节点且均存在于给定的二叉树中。</pre>\n",
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"difficulty": "Medium",
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"likes": 102,
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"dislikes": 0,
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"similarQuestions": "[]",
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"contributors": [],
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"langToValidPlayground": "{\"cpp\": true, \"java\": true, \"python\": true, \"python3\": true, \"mysql\": false, \"mssql\": false, \"oraclesql\": false, \"c\": false, \"csharp\": false, \"javascript\": false, \"typescript\": false, \"bash\": false, \"php\": false, \"swift\": false, \"kotlin\": false, \"dart\": false, \"golang\": false, \"ruby\": false, \"scala\": false, \"html\": false, \"pythonml\": false, \"rust\": false, \"racket\": false, \"erlang\": false, \"elixir\": false, \"pythondata\": false, \"react\": false, \"vanillajs\": false, \"postgresql\": false}",
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"topicTags": [
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{
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"name": "Tree",
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"slug": "tree",
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"translatedName": "树",
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"__typename": "TopicTagNode"
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},
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{
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"name": "Depth-First Search",
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"slug": "depth-first-search",
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"translatedName": "深度优先搜索",
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"__typename": "TopicTagNode"
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},
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{
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"name": "Binary Tree",
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"slug": "binary-tree",
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"translatedName": "二叉树",
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"__typename": "TopicTagNode"
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}
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"lang": "C++",
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"langSlug": "cpp",
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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 Solution {\npublic:\n TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {\n\n }\n};",
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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 TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {\n\n }\n}",
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"__typename": "CodeSnippetNode"
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"lang": "Python",
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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 Solution(object):\n def lowestCommonAncestor(self, root, p, q):\n \"\"\"\n :type root: TreeNode\n :type p: TreeNode\n :type q: TreeNode\n :rtype: TreeNode\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 lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:",
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"lang": "C",
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"langSlug": "c",
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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\nstruct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q){\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 TreeNode LowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {\n\n }\n}",
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"__typename": "CodeSnippetNode"
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},
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{
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"lang": "JavaScript",
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"langSlug": "javascript",
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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} root\n * @param {TreeNode} p\n * @param {TreeNode} q\n * @return {TreeNode}\n */\nvar lowestCommonAncestor = function(root, p, q) {\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 lowestCommonAncestor(root *TreeNode, p *TreeNode, q *TreeNode) *TreeNode {\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} root\n# @param {TreeNode} p\n# @param {TreeNode} q\n# @return {TreeNode}\ndef lowest_common_ancestor(root, p, q)\n\nend",
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"__typename": "CodeSnippetNode"
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"stats": "{\"totalAccepted\": \"29.3K\", \"totalSubmission\": \"41K\", \"totalAcceptedRaw\": 29318, \"totalSubmissionRaw\": 41028, \"acRate\": \"71.5%\"}",
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"hints": [
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"小心!你的算法处理只有一个节点的情况吗?会发生什么事?你可能要微调返回值。",
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"如果每个节点都有一个到其父节点的链接,我们可以利用9.2节问题2.7的方法。然而,面试官可能不会让我们作出这样的假设。",
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"第一个共同的祖先是最深的节点,这样p和q都是后代。想想你要如何识别这个节点。",
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"你如何弄清p是否为节点n的后代?",
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"从根节点开始。你能确定根是第一个共同祖先吗?如果不是,你能分辨出第一个共同祖先在根节点的哪一边吗?",
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"尝试递归方法。检查p和q是否为左子树和右子树的后代。如果它们是不同的树的后代,那么当前节点是第一个共同的祖先。如果它们是同一子树的后代,则该子树保存第一个共同祖先。现在,你该如何有效地实现它呢?",
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"在更简单的算法中,我们有一个方法表明x是n的后代,另一个方法是递归查找第一个共同的祖先。这样是在子树中反复搜索相同的元素。我们应该将其合并成一个firstCommonAncestor方法。那么什么样的返回值会给我们需要的信息?",
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"firstCommonAncestor函数可以返回第一个共同的祖先(如果p和q都包含在树里),如果p在树上而q不在,返回p;如果q在树上而p不在,返回q;否则,返回空。"
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],
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"solution": null,
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"status": null,
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"sampleTestCase": "[3,5,1,6,2,0,8,null,null,7,4]\n5\n1",
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