Flatten Binary Tree to Linked List
<p>Given the <code>root</code> of a binary tree, flatten the tree into a &quot;linked list&quot;:</p>

<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"><strong>pre-order</strong><strong> traversal</strong></a> of the binary tree.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0114.Flatten%20Binary%20Tree%20to%20Linked%20List/images/flaten.jpg" style="width: 500px; height: 226px;" />
<pre>
<strong>Input:</strong> root = [1,2,5,3,4,null,6]
<strong>Output:</strong> [1,null,2,null,3,null,4,null,5,null,6]
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<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>

<p>&nbsp;</p>
<strong>Follow up:</strong> Can you flatten the tree in-place (with <code>O(1)</code> extra space)?
Stack; Tree; Depth-First Search; Linked List; Binary Tree
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        while root:
            if root.left:
                pre = root.left
                while pre.right:
                    pre = pre.right
                pre.right = root.right
                root.right = root.left
                root.left = None
            root = root.right
