LeetCode 114: Flatten Binary Tree to Linked List — Python Solution

Solve LeetCode 114: Flatten Binary Tree to Linked List in Python with a reverse preorder threading approach. The key is to make the state invariant explicit, so the implementation and complexity follow naturally.

This guide paraphrases the task and does not reproduce LeetCode’s prompt. Use the official page for the complete statement, examples, constraints, and submission runner.

DifficultyMedium
TopicBinary Tree General
Reusable patternreverse preorder threading
ComplexityO(n) time and O(h) recursion space

What the problem is testing

Traverse right then left while keeping the previously processed node, and make each node point right to that previous node with left cleared.

Algorithm

  1. Traverse right then left while keeping the previously processed node, and make each node point right to that previous node with left cleared.
  2. Maintain this invariant: prev is the already flattened successor sequence for the current node in preorder.
  3. Continue until every input item or reachable state has been resolved, then return the accumulated result.

Python solution

LeetCode provides the list, tree, or graph node definition used by the method.

from collections import Counter, defaultdict, deque, OrderedDict
import random

class Solution:
    def flatten(self, root):
        previous = None
        def visit(node):
            nonlocal previous
            if not node:
                return
            visit(node.right)
            visit(node.left)
            node.right = previous
            node.left = None
            previous = node
        visit(root)

Why this is correct

The proof follows the maintained state: prev is the already flattened successor sequence for the current node in preorder. Each iteration preserves that claim while permanently resolving at least one position, node, interval, or search state. When the loop or recursion ends, every candidate required by the problem has therefore been included or ruled out, so the returned value is correct.

Complexity

O(n) time and O(h) recursion space. The stated auxiliary space excludes the returned output unless the output is the data structure being built.

Edge cases

Every left pointer must finish null.

Tested reference code

This implementation is included in the site’s downloadable 100-solution Python library. The complete suite compiles every solution and runs a behavioral assertion for every problem before publication.


Browse the searchable 100 LeetCode Python Solutions hub. Previous: 117. Populating Next Right Pointers in Each Node II · Next: 112. Path Sum