Leetcode #426: Convert Binary Search Tree to Sorted Doubly Linked List
In this guide, we solve Leetcode #426 Convert Binary Search Tree to Sorted Doubly 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.
Problem Statement
Convert a Binary Search Tree to a sorted Circular Doubly-Linked List in place. You can think of the left and right pointers as synonymous to the predecessor and successor pointers in a doubly-linked list.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Stack, Tree, Depth-First Search, Binary Search Tree, Linked List, Binary Tree, Doubly-Linked List
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 = [4,2,5,1,3]
Output: [1,2,3,4,5]
Explanation: The figure below shows the transformed BST. The solid line indicates the successor relationship, while the dashed line means the predecessor relationship.
Python Solution
"""
# Definition for a Node.
class Node:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
"""
class Solution:
def treeToDoublyList(self, root: 'Optional[Node]') -> 'Optional[Node]':
def dfs(root):
if root is None:
return
nonlocal prev, head
dfs(root.left)
if prev:
prev.right = root
root.left = prev
else:
head = root
prev = root
dfs(root.right)
if root is None:
return None
head = prev = None
dfs(root)
prev.right = head
head.left = prev
return head
Complexity
The time complexity is O(n). The space complexity is O(n).
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.