## DEV Community

Abhishek Chaudhary

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# Flatten Binary Tree to Linked List

Given the `root` of a binary tree, flatten the tree into a "linked list":

• The "linked list" should use the same `TreeNode` class where the `right` child pointer points to the next node in the list and the `left` child pointer is always `null`.
• The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [0]
Output: [0]

Constraints:

• The number of nodes in the tree is in the range `[0, 2000]`.
• `-100 <= Node.val <= 100`

Follow up: Can you flatten the tree in-place (with `O(1)` extra space)?SOLUTION:

``````# 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:
if root:
temp = root.right
root.right = self.flatten(root.left)
root.left = None
curr = root
while curr and curr.right:
curr = curr.right
curr.right = self.flatten(temp)
return root
``````