# 二叉树展开为链表

<p>给你二叉树的根结点 <code>root</code> ，请你将它展开为一个单链表：</p>

<ul>
	<li>展开后的单链表应该同样使用 <code>TreeNode</code> ，其中 <code>right</code> 子指针指向链表中下一个结点，而左子指针始终为 <code>null</code> 。</li>
	<li>展开后的单链表应该与二叉树 <a href="https://baike.baidu.com/item/%E5%85%88%E5%BA%8F%E9%81%8D%E5%8E%86/6442839?fr=aladdin" target="_blank"><strong>先序遍历</strong></a> 顺序相同。</li>
</ul>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/92075cab0821ae4b0edf18d3de9db0f1.jpeg" style="width: 500px; height: 226px;" />
<pre>
<strong>输入：</strong>root = [1,2,5,3,4,null,6]
<strong>输出：</strong>[1,null,2,null,3,null,4,null,5,null,6]
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>root = []
<strong>输出：</strong>[]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>root = [0]
<strong>输出：</strong>[0]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>树中结点数在范围 <code>[0, 2000]</code> 内</li>
	<li><code>-100 <= Node.val <= 100</code></li>
</ul>

<p> </p>

<p><strong>进阶：</strong>你可以使用原地算法（<code>O(1)</code> 额外空间）展开这棵树吗？</p>


## template

```cpp
#include <bits/stdc++.h>
using namespace std;

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution
{
public:
    void rconnect(TreeNode *&node, TreeNode *pmove)
    {
        if (pmove == nullptr)
            return;
        node->right = new TreeNode(pmove->val);
        node->left = nullptr;
        node = node->right;
        rconnect(node, pmove->left);
        rconnect(node, pmove->right);
    }
    void flatten(TreeNode *root)
    {
        if (root == nullptr)
            return;
        TreeNode *head = new TreeNode();
        TreeNode *newroot = head;
        rconnect(head, root);
        newroot = newroot->right->right;
        root->right = newroot;
        root->left = nullptr;
    }
};

```

## 答案

```cpp

```

## 选项

### A

```cpp

```

### B

```cpp

```

### C

```cpp

```