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Problem solution · Python

Convert Binary Search Tree to Sorted Doubly Linked List

Convert Binary Search Tree to Sorted Doubly Linked List: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
26 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Convert Binary Search Tree to Sorted Doubly Linked List, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 26 lines of Python from the credited upstream file 426-2.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeConvert Binary Search Tree to Sorted Doubly Linked List · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def treeToDoublyList(self, root: 'Node | None') -> 'Node | None':    if not root:      return None     stack = []    first = None    pred = None     while root or stack:      while root:        stack.append(root)        root = root.left      root = stack.pop()      if not first:        first = root      if pred:        pred.right = root        root.left = pred      pred = root      root = root.right     pred.right = first    first.left = pred    return first

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