Fetch the repository succeeded.
This action will force synchronization from 小墨/力扣题库(完整版), which will overwrite any changes that you have made since you forked the repository, and can not be recovered!!!
Synchronous operation will process in the background and will refresh the page when finishing processing. Please be patient.
<p>You are given a tree with <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code> in the form of a parent array <code>parent</code> where <code>parent[i]</code> is the parent of <code>i<sup>th</sup></code> node. The root of the tree is node <code>0</code>. Find the <code>k<sup>th</sup></code> ancestor of a given node.</p>
<p>The <code>k<sup>th</sup></code> ancestor of a tree node is the <code>k<sup>th</sup></code> node in the path from that node to the root node.</p>
<p>Implement the <code>TreeAncestor</code> class:</p>
<ul>
<li><code>TreeAncestor(int n, int[] parent)</code> Initializes the object with the number of nodes in the tree and the parent array.</li>
<li><code>int getKthAncestor(int node, int k)</code> return the <code>k<sup>th</sup></code> ancestor of the given node <code>node</code>. If there is no such ancestor, return <code>-1</code>.</li>
</ul>
<p> </p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/08/28/1528_ex1.png" style="width: 396px; height: 262px;" />
<pre>
<strong>Input</strong>
["TreeAncestor", "getKthAncestor", "getKthAncestor", "getKthAncestor"]
[[7, [-1, 0, 0, 1, 1, 2, 2]], [3, 1], [5, 2], [6, 3]]
<strong>Output</strong>
[null, 1, 0, -1]
<strong>Explanation</strong>
TreeAncestor treeAncestor = new TreeAncestor(7, [-1, 0, 0, 1, 1, 2, 2]);
treeAncestor.getKthAncestor(3, 1); // returns 1 which is the parent of 3
treeAncestor.getKthAncestor(5, 2); // returns 0 which is the grandparent of 5
treeAncestor.getKthAncestor(6, 3); // returns -1 because there is no such ancestor</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= k <= n <= 5 * 10<sup>4</sup></code></li>
<li><code>parent.length == n</code></li>
<li><code>parent[0] == -1</code></li>
<li><code>0 <= parent[i] < n</code> for all <code>0 < i < n</code></li>
<li><code>0 <= node < n</code></li>
<li>There will be at most <code>5 * 10<sup>4</sup></code> queries.</li>
</ul>
此处可能存在不合适展示的内容,页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容,可点击提交进行申诉,我们将尽快为您处理。