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Completed Trees-1#1750

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Completed Trees-1#1750
Sanket-S-Kale wants to merge 1 commit into
super30admin:masterfrom
Sanket-S-Kale:master

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Validate Binary Search Tree (Problem1.py)

Strengths:

  1. Excellent use of boundary tracking approach - more intuitive and often more efficient than the in-order traversal method
  2. Clear, well-documented code with helpful comments explaining the logic
  3. Proper handling of edge cases (None boundaries)
  4. Good variable naming that makes the code self-documenting
  5. The commented-out in-order traversal approach shows awareness of multiple solution strategies

Areas for Improvement:

  1. The time complexity analysis comment mentions "O(N)" but doesn't specify what N represents (should clarify "where N is the number of nodes")
  2. Consider adding type hints for better code documentation (e.g., def helper(node: Optional[TreeNode], minVal: Optional[int], maxVal: Optional[int]) -> bool:)
  3. The solution could benefit from early termination optimization - currently it continues checking even after finding an invalid node (though this is minor since Python's and operator short-circuits)

Minor Suggestions:

  • The boundary check if minVal is not None and node.val <= minVal could be simplified to if node.val <= minVal since comparisons with None work correctly in Python, but the current approach is more explicit and readable

VERDICT: PASS


Construct Binary Tree from Preorder and Inorder Traversal (Problem2.py)

Strengths:

  • Clean, readable code with excellent documentation
  • Correct algorithmic approach
  • Good variable naming conventions
  • Proper handling of edge cases

Areas for Improvement:

  1. Use a hash map for index lookups: Instead of searching through inorder linearly for each node, store the inorder indices in a dictionary for O(1) access. This is the key optimization that reduces time from O(N²) to O(N).

  2. Use index ranges instead of list slicing: Instead of creating new arrays via slicing (which is O(N) and consumes extra memory), pass start and end indices to the recursive function. This reduces both time and space complexity.

  3. Consider using an index variable for preorder: Instead of slicing preorder, use a global/index parameter to track the current position, similar to how the reference solution uses idx.

Suggested Optimization Pattern:

# Instead of:
inorderRootIdx = -1
for i in range(len(inorder)):
    if inorder[i] == rootVal:
        inorderRootIdx = i
        break

# Use:
map = {val: idx for idx, val in enumerate(inorder)}
inorderRootIdx = map[rootVal]

VERDICT: NEEDS_IMPROVEMENT

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