Flatten Binary Tree to Linked List

Tested Python solution for LeetCode 114 with 16 pytest cases. Generate a practice environment with lcpy.

LeetCode 114, Medium. Topics: Linked List, Stack, Tree, Depth-First Search, Binary Tree. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 16 parametrized pytest cases, and a playground notebook:

lcpy gen -n 114   # by problem number
lcpy gen -s flatten_binary_tree_to_linked_list   # by problem name

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Problem

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.

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Examples

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

Input: root = []
Output: []

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

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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)?

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

from leetcode_py import TreeNode


class Solution:
    # Time: O(n), each node is visited a constant number of times
    # Space: O(1), pointers are rewired in place
    def flatten(self, root: TreeNode[int] | None) -> None:
        current = root
        while current is not None:
            if current.left is not None:
                predecessor = current.left
                while predecessor.right is not None:
                    predecessor = predecessor.right
                predecessor.right = current.right
                current.right = current.left
                current.left = None
            current = current.right

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Complexity

Time	Space
O(n), each node is visited a constant number of times	O(1), pointers are rewired in place

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