# 114-flatten-binary-tree-to-linked-list


Try it on <a href='https://leetcode.com/problems/114-flatten-binary-tree-to-linked-list'>leetcode</a>

## Description
<div class="description">
<div><p>Given the <code>root</code> of a binary tree, flatten the tree into a "linked list":</p>

<ul>
	<li>The "linked list" 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 "linked list" 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>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/14/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>Example 2:</strong></p>

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

<p><strong>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)?</div>
</div>

## Solution(Python)
```Python
# 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.
        """
        cur = root
        while cur:
            if cur.left:
                pre = cur.left
                while pre.right:
                    pre = pre.right
                
                pre.right = cur.right
                
                cur.right = cur.left
                cur.left = None
            cur = cur.right 
```