1
0
mirror of https://gitee.com/coder-xiaomo/leetcode-problemset synced 2025-01-10 18:48:13 +08:00
Code Issues Projects Releases Wiki Activity GitHub Gitee
leetcode-problemset/leetcode/problem/lowest-common-ancestor-of-a-binary-tree.html
2022-03-29 12:55:24 +08:00

39 lines
1.7 KiB
HTML

<p>Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.</p>
<p>According to the <a href="https://en.wikipedia.org/wiki/Lowest_common_ancestor" target="_blank">definition of LCA on Wikipedia</a>: &ldquo;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>).&rdquo;</p>
<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong> The LCA of nodes 5 and 1 is 3.
</pre>
<p><strong>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
<strong>Output:</strong> 5
<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.
</pre>
<p><strong>Example 3:</strong></p>
<pre>
<strong>Input:</strong> root = [1,2], p = 1, q = 2
<strong>Output:</strong> 1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li>The number of nodes in the tree is in the range <code>[2, 10<sup>5</sup>]</code>.</li>
<li><code>-10<sup>9</sup> &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
<li>All <code>Node.val</code> are <strong>unique</strong>.</li>
<li><code>p != q</code></li>
<li><code>p</code> and <code>q</code> will exist in the tree.</li>
</ul>