mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
37 lines
1.5 KiB
HTML
37 lines
1.5 KiB
HTML
|
<p>Given the <code>root</code> of a binary search tree (BST) with duplicates, return <em>all the <a href="https://en.wikipedia.org/wiki/Mode_(statistics)" target="_blank">mode(s)</a> (i.e., the most frequently occurred element) in it</em>.</p>
|
||
|
|
|
||
|
|
<p>If the tree has more than one mode, return them in <strong>any order</strong>.</p>
|
||
|
|
|
||
|
|
<p>Assume a BST is defined as follows:</p>
|
||
|
|
|
||
|
|
<ul>
|
||
|
|
<li>The left subtree of a node contains only nodes with keys <strong>less than or equal to</strong> the node's key.</li>
|
||
|
|
<li>The right subtree of a node contains only nodes with keys <strong>greater than or equal to</strong> the node's key.</li>
|
||
|
|
<li>Both the left and right subtrees must also be binary search trees.</li>
|
||
|
|
</ul>
|
||
|
|
|
||
|
|
<p> </p>
|
||
|
|
<p><strong>Example 1:</strong></p>
|
||
|
|
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/11/mode-tree.jpg" style="width: 142px; height: 222px;" />
|
||
|
|
<pre>
|
||
|
|
<strong>Input:</strong> root = [1,null,2,2]
|
||
|
|
<strong>Output:</strong> [2]
|
||
|
|
</pre>
|
||
|
|
|
||
|
|
<p><strong>Example 2:</strong></p>
|
||
|
|
|
||
|
|
<pre>
|
||
|
|
<strong>Input:</strong> root = [0]
|
||
|
|
<strong>Output:</strong> [0]
|
||
|
|
</pre>
|
||
|
|
|
||
|
|
<p> </p>
|
||
|
|
<p><strong>Constraints:</strong></p>
|
||
|
|
|
||
|
|
<ul>
|
||
|
|
<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
|
||
|
|
<li><code>-10<sup>5</sup> <= Node.val <= 10<sup>5</sup></code></li>
|
||
|
|
</ul>
|
||
|
|
|
||
|
|
<p> </p>
|
||
|
|
<strong>Follow up:</strong> Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).
|