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Problem solution · Python

Maximum Sum BST in Binary Tree

Maximum Sum BST in Binary Tree: 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
35 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 Maximum Sum BST in Binary Tree, 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

  • 35 lines of Python from the credited upstream file 1373.py.
  • The implementation keeps its working state in language-native values and containers.
  • 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 codeMaximum Sum BST in Binary Tree · PythonPython
Use this to learn the idea, then write your own version.
from dataclasses import dataclass  @dataclassclass T:  isBST: bool | None = False  mx: int | None = None  mn: int | None = None  summ: int | None = None  class Solution:  def maxSumBST(self, root: TreeNode | None) -> int:    self.ans = 0     def traverse(root: TreeNode | None) -> T:      if not root:        return T(True, -math.inf, math.inf, 0)       left: T = traverse(root.left)      right: T = traverse(root.right)       if not left.isBST or not right.isBST:        return T()      if root.val <= left.mx or root.val >= right.mn:        return T()       # The `root` is a valid BST.      summ = root.val + left.summ + right.summ      self.ans = max(self.ans, summ)      return T(True, max(root.val, right.mx), min(root.val, left.mn), summ)     traverse(root)    return self.ans

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