27 lines
1.1 KiB
Markdown
27 lines
1.1 KiB
Markdown
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<p>给你一棵二叉搜索树，请你 <strong>按中序遍历</strong> 将其重新排列为一棵递增顺序搜索树，使树中最左边的节点成为树的根节点，并且每个节点没有左子节点，只有一个右子节点。</p>
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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/2020/11/17/ex1.jpg" style="width: 600px; height: 350px;" />
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<pre>
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<strong>输入：</strong>root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
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<strong>输出：</strong>[1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
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</pre>
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<p><strong>示例 2：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/11/17/ex2.jpg" style="width: 300px; height: 114px;" />
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<pre>
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<strong>输入：</strong>root = [5,1,7]
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<strong>输出：</strong>[1,null,5,null,7]
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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>树中节点数的取值范围是 <code>[1, 100]</code></li>
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<li><code>0 <= Node.val <= 1000</code></li>
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</ul>
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<div><div>Related Topics</div><div><li>树</li><li>深度优先搜索</li><li>递归</li></div></div>\n<div><li>👍 154</li><li>👎 0</li></div>
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