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48 lines
1.3 KiB
HTML
48 lines
1.3 KiB
HTML
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<p>给定一个二叉树</p>
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<pre>
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struct Node {
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int val;
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Node *left;
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Node *right;
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Node *next;
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}</pre>
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<p>填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 <code>NULL</code>。</p>
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<p>初始状态下，所有 next 指针都被设置为 <code>NULL</code>。</p>
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<p> </p>
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<p><strong>进阶：</strong></p>
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<ul>
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<li>你只能使用常量级额外空间。</li>
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<li>使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。</li>
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</ul>
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<p> </p>
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<p><strong>示例：</strong></p>
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<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/02/15/117_sample.png" style="height: 218px; width: 640px;" /></p>
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<pre>
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<strong>输入</strong>：root = [1,2,3,4,5,null,7]
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<strong>输出：</strong>[1,#,2,3,#,4,5,7,#]
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<strong>解释：</strong>给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。序列化输出按层序遍历顺序（由 next 指针连接），'#' 表示每层的末尾。</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>树中的节点数小于 <code>6000</code></li>
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<li><code>-100 <= node.val <= 100</code></li>
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</ul>
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<p> </p>
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<ul>
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</ul>
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