flatten-binary-tree-to-linked-list

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Algorithms

Given the <code>root</code> of a binary tree, flatten the tree into a &quot;linked list&quot;:

<ul>
	<li>The &quot;linked list&quot; should use the same <code>TreeNode</code> class where the <code>right</code> child pointer points to the next node in the list and the <code>left</code> child pointer is always <code>null</code>.</li>
	<li>The &quot;linked list&quot; should be in the same order as a <a href="https://en.wikipedia.org/wiki/Tree_traversal#Pre-order,_NLR" target="_blank">pre-order traversal</a> of the binary tree.</li>
</ul>

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/14/flaten.jpg" style="width: 500px; height: 226px;" />
<pre>
Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]
</pre>

Example 2:

<pre>
Input: root = []
Output: []
</pre>

Example 3:

<pre>
Input: root = [0]
Output: [0]
</pre>

&nbsp;
Constraints:

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 2000]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

&nbsp;
Follow up: Can you flatten the tree in-place (with <code>O(1)</code> extra space)?

flatten

Flatten Binary Tree to Linked List

Flatten a Multilevel Doubly Linked List

flatten-a-multilevel-doubly-linked-list

扁平化多级双向链表

Correct a Binary Tree

correct-a-binary-tree

纠正二叉树

Stack

Tree

Depth-First Search

Linked List

Binary Tree

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

<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">先序遍历</a> 顺序相同。</li>
</ul>

 

示例 1：
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/14/flaten.jpg" style="width: 500px; height: 226px;" />
<pre>
输入：root = [1,2,5,3,4,null,6]
输出：[1,null,2,null,3,null,4,null,5,null,6]
</pre>

示例 2：

<pre>
输入：root = []
输出：[]
</pre>

示例 3：

<pre>
输入：root = [0]
输出：[0]
</pre>

 

提示：

<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>

 

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