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34 lines
1.2 KiB
HTML
34 lines
1.2 KiB
HTML
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<p>给定一个二叉树，我们在树的节点上安装摄像头。</p>
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<p>节点上的每个摄影头都可以监视<strong>其父对象、自身及其直接子对象。</strong></p>
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<p>计算监控树的所有节点所需的最小摄像头数量。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/29/bst_cameras_01.png" style="height: 163px; width: 138px;"></p>
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<pre><strong>输入：</strong>[0,0,null,0,0]
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<strong>输出：</strong>1
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<strong>解释：</strong>如图所示，一台摄像头足以监控所有节点。
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</pre>
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<p><strong>示例 2：</strong></p>
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<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/29/bst_cameras_02.png" style="height: 312px; width: 139px;"></p>
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<pre><strong>输入：</strong>[0,0,null,0,null,0,null,null,0]
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<strong>输出：</strong>2
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<strong>解释：</strong>需要至少两个摄像头来监视树的所有节点。 上图显示了摄像头放置的有效位置之一。
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</pre>
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<p><br>
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<strong>提示：</strong></p>
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<ol>
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<li>给定树的节点数的范围是 <code>[1, 1000]</code>。</li>
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<li>每个节点的值都是 0。</li>
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</ol>
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