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41 lines
1.4 KiB
HTML
41 lines
1.4 KiB
HTML
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<p>给你一个含重复值的二叉搜索树（BST）的根节点 <code>root</code> ，找出并返回 BST 中的所有 <a href="https://baike.baidu.com/item/%E4%BC%97%E6%95%B0/44796" target="_blank">众数</a>（即，出现频率最高的元素）。</p>
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<p>如果树中有不止一个众数，可以按 <strong>任意顺序</strong> 返回。</p>
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<p>假定 BST 满足如下定义：</p>
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<ul>
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<li>结点左子树中所含节点的值 <strong>小于等于</strong> 当前节点的值</li>
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<li>结点右子树中所含节点的值 <strong>大于等于</strong> 当前节点的值</li>
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<li>左子树和右子树都是二叉搜索树</li>
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</ul>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/03/11/mode-tree.jpg" style="width: 142px; height: 222px;" />
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<pre>
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<strong>输入：</strong>root = [1,null,2,2]
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<strong>输出：</strong>[2]
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>root = [0]
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<strong>输出：</strong>[0]
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li>树中节点的数目在范围 <code>[1, 10<sup>4</sup>]</code> 内</li>
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<li><code>-10<sup>5</sup> <= Node.val <= 10<sup>5</sup></code></li>
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</ul>
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<p> </p>
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<p><strong>进阶：</strong>你可以不使用额外的空间吗？（假设由递归产生的隐式调用栈的开销不被计算在内）</p>
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