leetcode-problemset/leetcode-cn/originData/first-common-ancestor-lcci.json

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            "questionId": "100179",
            "questionFrontendId": "面试题 04.08",
            "categoryTitle": "LCCI",
            "boundTopicId": 46238,
            "title": "First Common Ancestor LCCI",
            "titleSlug": "first-common-ancestor-lcci",
            "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",
            "translatedTitle": "首个共同祖先",
            "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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            "topicTags": [
                {
                    "name": "Tree",
                    "slug": "tree",
                    "translatedName": "树",
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                    "name": "Depth-First Search",
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                    "translatedName": "深度优先搜索",
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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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            "hints": [
                "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。",
                "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。",
                "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。",
                "你如何弄清p是否为节点n的后代？",
                "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？",
                "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？",
                "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？",
                "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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