update

leetcode-cn/originData/lowest-common-ancestor-of-deepest-leaves.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个有根节点<meta charset=\"UTF-8\" />&nbsp;<code>root</code>&nbsp;的二叉树，返回它&nbsp;<em>最深的叶节点的最近公共祖先</em>&nbsp;。</p>\n\n<p>回想一下：</p>\n\n<ul>\n\t<li><strong>叶节点</strong> 是二叉树中没有子节点的节点</li>\n\t<li>树的根节点的&nbsp;<strong>深度&nbsp;</strong>为&nbsp;<code>0</code>，如果某一节点的深度为&nbsp;<code>d</code>，那它的子节点的深度就是&nbsp;<code>d+1</code></li>\n\t<li>如果我们假定 <code>A</code> 是一组节点&nbsp;<code>S</code>&nbsp;的 <strong>最近公共祖先</strong>，<code>S</code>&nbsp;中的每个节点都在以 <code>A</code> 为根节点的子树中，且 <code>A</code>&nbsp;的深度达到此条件下可能的最大值。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/01/sketch1.png\" style=\"height: 340px; width: 400px;\" />\n<pre>\n<strong>输入：</strong>root = [3,5,1,6,2,0,8,null,null,7,4]\n<strong>输出：</strong>[2,7,4]\n<strong>解释：</strong>我们返回值为 2 的节点，在图中用黄色标记。\n在图中用蓝色标记的是树的最深的节点。\n注意，节点 6、0 和 8 也是叶节点，但是它们的深度是 2 ，而节点 7 和 4 的深度是 3 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [1]\n<strong>输出：</strong>[1]\n<strong>解释：</strong>根节点是树中最深的节点，它是它本身的最近公共祖先。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [0,1,3,null,2]\n<strong>输出：</strong>[2]\n<strong>解释：</strong>树中最深的叶节点是 2 ，最近公共祖先是它自己。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中的节点数将在<meta charset=\"UTF-8\" />&nbsp;<code>[1, 1000]</code>&nbsp;的范围内。</li>\n\t<li><code>0 &lt;= Node.val &lt;= 1000</code></li>\n\t<li>每个节点的值都是&nbsp;<strong>独一无二</strong>&nbsp;的。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>注意：</strong>本题与力扣 865 重复：<a href=\"https://leetcode-cn.com/problems/smallest-subtree-with-all-the-deepest-nodes/\">https://leetcode-cn.com/problems/smallest-subtree-with-all-the-deepest-nodes/</a></p>\n",
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             "difficulty": "Medium",
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@@ -161,7 +161,7 @@
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             "hints": [
                 "Do a postorder traversal.",
                 "Then, if both subtrees contain a deepest leaf, you can mark this node as the answer (so far).",