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Tree1#1747

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YogeshPardeshi wants to merge 2 commits into
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Tree1#1747
YogeshPardeshi wants to merge 2 commits into
super30admin:masterfrom
YogeshPardeshi:master

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@super30admin
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Validate BST (Problem1.java)

Strengths:

  1. Correct implementation of the in-order traversal BST validation approach
  2. Clean and readable code structure
  3. Proper use of instance variables to track state across recursive calls
  4. Good variable naming (flag, prev)

Areas for Improvement:

  1. Class name should follow standard conventions (e.g., "Solution" instead of "Problem1")
  2. Consider adding comments explaining the in-order traversal approach for BST validation
  3. The solution could benefit from early termination when flag becomes false (though this is a minor optimization)

The solution is functionally equivalent to the reference solution and demonstrates a solid understanding of BST properties and in-order traversal.

VERDICT: PASS


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

Strengths:

  • Clean, readable code with meaningful variable names
  • Correct recursive logic for tree construction
  • Good use of Java's Arrays.copyOfRange for array splitting

Areas for Improvement:

  1. Performance Issue: The linear search for rootIdx in inorder at each recursive call is inefficient. Use a HashMap to store value-to-index mappings like the reference solution (O(n) preprocessing, O(1) lookups).

  2. Array Copying Overhead: Creating new arrays at each recursive level is memory-intensive. Consider using indices (start/end pointers) instead of array copying to improve both time and space efficiency.

  3. Base Case: The base case preorder.length == 0 is not robust. Use if (start > end) return null; to properly handle empty segments.

  4. Missing HashMap Optimization: The reference solution achieves O(n) time complexity using a HashMap. This is a significant optimization opportunity.

Suggested Improvement:

// Use a HashMap for O(1) lookups and indices instead of array copying
HashMap<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < inorder.length; i++) {
    map.put(inorder[i], i);
}
return build(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, map);

VERDICT: NEEDS_IMPROVEMENT

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2 participants