Leetcode #114: Flatten Binary Tree to Linked List

In this guide, we solve Leetcode #114 Flatten Binary Tree to Linked List in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

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.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Stack, Tree, Depth-First Search, Linked List, Binary Tree

Intuition

The problem has a natural nested or last-in-first-out structure.

A stack lets us resolve matches in the correct order as we scan.

Approach

Push items as they appear and pop when you can finalize a decision.

The stack captures the unresolved part of the input.

Steps:

Push elements as you scan.
Pop when a rule or match is satisfied.
Use the stack to compute results.

Example

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

Python 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.
        """
        while root:
            if root.left:
                pre = root.left
                while pre.right:
                    pre = pre.right
                pre.right = root.right
                root.right = root.left
                root.left = None
            root = root.right

Complexity

The time complexity is $O(n)$ , where $n$ is the number of nodes in the tree. The space complexity is $O(1)$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Problem Statement

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.

Python 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.
        """
        while root:
            if root.left:
                pre = root.left
                while pre.right:
                    pre = pre.right
                pre.right = root.right
                root.right = root.left
                root.left = None
            root = root.right

Leetcode #114: Flatten Binary Tree to Linked List

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #114: Flatten Binary Tree to Linked List

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview