update

leetcode-cn/originData/kth-ancestor-of-a-tree-node.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一棵树，树上有 <code>n</code> 个节点，按从 <code>0</code> 到 <code>n-1</code> 编号。树以父节点数组的形式给出，其中 <code>parent[i]</code> 是节点 <code>i</code> 的父节点。树的根节点是编号为 <code>0</code> 的节点。</p>\n\n<p>树节点的第 <em><code>k</code> </em>个祖先节点是从该节点到根节点路径上的第 <code>k</code> 个节点。</p>\n\n<p>实现 <code>TreeAncestor</code> 类：</p>\n\n<ul>\n\t<li><code>TreeAncestor（int n， int[] parent）</code> 对树和父数组中的节点数初始化对象。</li>\n\t<li><code>getKthAncestor</code><code>(int node, int k)</code> 返回节点 <code>node</code> 的第 <code>k</code> 个祖先节点。如果不存在这样的祖先节点，返回 <code>-1</code>&nbsp;。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/06/14/1528_ex1.png\" /></strong></p>\n\n<pre>\n<strong>输入：</strong>\n[\"TreeAncestor\",\"getKthAncestor\",\"getKthAncestor\",\"getKthAncestor\"]\n[[7,[-1,0,0,1,1,2,2]],[3,1],[5,2],[6,3]]\n\n<strong>输出：</strong>\n[null,1,0,-1]\n\n<strong>解释：</strong>\nTreeAncestor treeAncestor = new TreeAncestor(7, [-1, 0, 0, 1, 1, 2, 2]);\n\ntreeAncestor.getKthAncestor(3, 1);  // 返回 1 ，它是 3 的父节点\ntreeAncestor.getKthAncestor(5, 2);  // 返回 0 ，它是 5 的祖父节点\ntreeAncestor.getKthAncestor(6, 3);  // 返回 -1 因为不存在满足要求的祖先节点\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>\n\t<li><code>parent[0] == -1</code> 表示编号为 <code>0</code> 的节点是根节点。</li>\n\t<li>对于所有的 <code>0 &lt;&nbsp;i &lt; n</code> ，<code>0 &lt;= parent[i] &lt; n</code> 总成立</li>\n\t<li><code>0 &lt;= node &lt; n</code></li>\n\t<li>至多查询&nbsp;<code>5 * 10<sup>4</sup></code> 次</li>\n</ul>\n",
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             "difficulty": "Hard",
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             "hints": [
                 "The queries must be answered efficiently to avoid time limit exceeded verdict.",
                 "Use sparse table (dynamic programming application) to travel the tree upwards in a fast way."