mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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156 lines
25 KiB
JSON
156 lines
25 KiB
JSON
{
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"data": {
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"question": {
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"questionId": "236",
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"questionFrontendId": "236",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1440,
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"title": "Lowest Common Ancestor of a Binary Tree",
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"titleSlug": "lowest-common-ancestor-of-a-binary-tree",
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"content": "<p>Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.</p>\n\n<p>According to the <a href=\"https://en.wikipedia.org/wiki/Lowest_common_ancestor\" target=\"_blank\">definition of LCA on Wikipedia</a>: “The lowest common ancestor is defined between two nodes <code>p</code> and <code>q</code> as the lowest node in <code>T</code> that has both <code>p</code> and <code>q</code> as descendants (where we allow <b>a node to be a descendant of itself</b>).”</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2018/12/14/binarytree.png\" style=\"width: 200px; height: 190px;\" />\n<pre>\n<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> The LCA of nodes 5 and 1 is 3.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2018/12/14/binarytree.png\" style=\"width: 200px; height: 190px;\" />\n<pre>\n<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [1,2], p = 1, q = 2\n<strong>Output:</strong> 1\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the tree is in the range <code>[2, 10<sup>5</sup>]</code>.</li>\n\t<li><code>-10<sup>9</sup> <= Node.val <= 10<sup>9</sup></code></li>\n\t<li>All <code>Node.val</code> are <strong>unique</strong>.</li>\n\t<li><code>p != q</code></li>\n\t<li><code>p</code> and <code>q</code> will exist in the tree.</li>\n</ul>\n",
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"translatedTitle": "二叉树的最近公共祖先",
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"translatedContent": "<p>给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。</p>\n\n<p><a href=\"https://baike.baidu.com/item/%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88/8918834?fr=aladdin\" target=\"_blank\">百度百科</a>中最近公共祖先的定义为：“对于有根树 T 的两个节点 p、q，最近公共祖先表示为一个节点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（<strong>一个节点也可以是它自己的祖先</strong>）。”</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2018/12/14/binarytree.png\" style=\"width: 200px; height: 190px;\" />\n<pre>\n<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 <code>5 </code>和节点 <code>1 </code>的最近公共祖先是节点 <code>3 。</code>\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2018/12/14/binarytree.png\" style=\"width: 200px; height: 190px;\" />\n<pre>\n<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 <code>5 </code>和节点 <code>4 </code>的最近公共祖先是节点 <code>5 。</code>因为根据定义最近公共祖先节点可以为节点本身。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [1,2], p = 1, q = 2\n<strong>输出：</strong>1\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中节点数目在范围 <code>[2, 10<sup>5</sup>]</code> 内。</li>\n\t<li><code>-10<sup>9</sup> <= Node.val <= 10<sup>9</sup></code></li>\n\t<li>所有 <code>Node.val</code> <code>互不相同</code> 。</li>\n\t<li><code>p != q</code></li>\n\t<li><code>p</code> 和 <code>q</code> 均存在于给定的二叉树中。</li>\n</ul>\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 Solution {\n func lowestCommonAncestor(_ root: TreeNode?, _ p: TreeNode?, _ q: TreeNode?) -> TreeNode? {\n \n }\n}",
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