Project Notes

Recover Binary Search Tree

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Description

tab: English
 
<p>You are given the <code>root</code> of a binary search tree (BST), where the values of <strong>exactly</strong> two nodes of the tree were swapped by mistake. <em>Recover the tree without changing its structure</em>.</p>
 
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/28/recover1.jpg" style="width: 422px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [1,3,null,null,2]
<strong>Output:</strong> [3,1,null,null,2]
<strong>Explanation:</strong> 3 cannot be a left child of 1 because 3 &gt; 1. Swapping 1 and 3 makes the BST valid.
</pre>
 
<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/28/recover2.jpg" style="width: 581px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [3,1,4,null,null,2]
<strong>Output:</strong> [2,1,4,null,null,3]
<strong>Explanation:</strong> 2 cannot be in the right subtree of 3 because 2 &lt; 3. Swapping 2 and 3 makes the BST valid.
</pre>
 
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
 
<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 1000]</code>.</li>
	<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
 
<p>&nbsp;</p>
<strong>Follow up:</strong> A solution using <code>O(n)</code> space is pretty straight-forward. Could you devise a constant <code>O(1)</code> space solution?
 
 
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[submissions](https://leetcode.com/problems/recover-binary-search-tree/submissions/) | [solutions](https://leetcode.com/problems/recover-binary-search-tree/solutions/)
 
 
tab: 中文
 
<p>给你二叉搜索树的根节点 <code>root</code> ，该树中的 <strong>恰好</strong> 两个节点的值被错误地交换。<em>请在不改变其结构的情况下，恢复这棵树&nbsp;</em>。</p>
 
<p>&nbsp;</p>
 
<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/28/recover1.jpg" style="width: 300px;" />
<pre>
<strong>输入：</strong>root = [1,3,null,null,2]
<strong>输出：</strong>[3,1,null,null,2]
<strong>解释：</strong>3 不能是 1 的左孩子，因为 3 &gt; 1 。交换 1 和 3 使二叉搜索树有效。
</pre>
 
<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/28/recover2.jpg" style="height: 208px; width: 400px;" />
<pre>
<strong>输入：</strong>root = [3,1,4,null,null,2]
<strong>输出：</strong>[2,1,4,null,null,3]
<strong>解释：</strong>2 不能在 3 的右子树中，因为 2 &lt; 3 。交换 2 和 3 使二叉搜索树有效。</pre>
 
<p>&nbsp;</p>
 
<p><strong>提示：</strong></p>
 
<ul>
	<li>树上节点的数目在范围 <code>[2, 1000]</code> 内</li>
	<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
 
<p>&nbsp;</p>
 
<p><strong>进阶：</strong>使用 <code>O(n)</code> 空间复杂度的解法很容易实现。你能想出一个只使用&nbsp;<code>O(1)</code> 空间的解决方案吗？</p>
 
 
 
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[提交记录](https://leetcode.cn/problems/recover-binary-search-tree/submissions/) | [题解](https://leetcode.cn/problems/recover-binary-search-tree/solution/)

Solutions & Notes

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