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leetcode-problemset/leetcode-cn/problem (Chinese)/翻转等价二叉树 [flip-equivalent-binary-trees].html

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<p>我们可以为二叉树 <strong>T</strong> 定义一个&nbsp;<strong>翻转操作&nbsp;</strong>,如下所示:选择任意节点,然后交换它的左子树和右子树。</p>
<p>只要经过一定次数的翻转操作后,能使 <strong>X</strong> 等于 <strong>Y</strong>,我们就称二叉树 <strong>X</strong> <em>翻转 等价&nbsp;</em>于二叉树 <strong>Y</strong></p>
<p>这些树由根节点&nbsp;<code>root1</code><code>root2</code>&nbsp;给出。如果两个二叉树是否是<em>翻转 等价&nbsp;</em>的函数,则返回 <code>true</code> ,否则返回 <code>false</code></p>
<p>&nbsp;</p>
<p><strong>示例 1</strong></p>
<p><img alt="Flipped Trees Diagram" src="https://assets.leetcode.com/uploads/2018/11/29/tree_ex.png" /></p>
<pre>
<strong>输入:</strong>root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
<strong>输出:</strong>true
<strong>解释:</strong>我们翻转值为 13 以及 5 的三个节点。
</pre>
<p><strong>示例&nbsp;2:</strong></p>
<pre>
<strong>输入:</strong> root1 = [], root2 = []
<strong>输出:</strong> true
</pre>
<p><strong>示例 3:</strong></p>
<pre>
<strong>输入:</strong> root1 = [], root2 = [1]
<strong>输出:</strong> false
</pre>
<p>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li>每棵树节点数在&nbsp;<code>[0, 100]</code> 范围内</li>
<li>每棵树中的每个值都是唯一的、在 <code>[0, 99]</code>&nbsp;范围内的整数</li>
</ul>