LeetCode - 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

 

My 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:
        """
        Do not return anything, modify root in-place instead.
        """
        
        if not root or not root.left and not root.right:
            return root
        
        right_node = root.right
        root.right = self.flatten(root.left)
        root.left = None

        if right_node:
            node = root
            while node.right:
                node = node.right
            node.right = self.flatten(right_node)

        return root