Completed Trees 3

[Keerthi0910]

No description provided.

[super30admin]

Path Sum II (Problem1.java)

Strengths:

1. Your solution correctly implements the DFS approach with backtracking.
2. The backtracking step (path.remove(path.size()-1)) is correctly placed after both recursive calls, which is the proper way to handle this.
3. The code is clean, readable, and follows Java conventions.
4. Using new ArrayList<>(path) to create a copy when adding to the result is the correct approach.

Areas for Improvement:

1. Complexity Analysis: Your claimed time complexity of O(n) is incorrect. While the traversal is O(n), copying paths when adding to results takes O(k) where k is the path length. In the worst case (skewed tree), this results in O(n²) time complexity. Similarly, your claimed space complexity of O(1) is misleading - you need O(h) space for the recursion stack and the current path, where h is the height of the tree.

2. Code Formatting: There's an extra blank line in your code that could be removed for cleaner formatting.

3. Minor Optimization: Consider adding a null check optimization at the start of helper to avoid unnecessary recursive calls, though this is a minor point.

VERDICT: PASS

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Symmetric Tree (Problem2.java)

Strengths:

• Correct and elegant recursive solution
• Clean code structure with helper method
• Proper handling of edge cases (null nodes)
• Good use of short-circuit evaluation

Areas for Improvement:

• Space complexity claim is incorrect: You stated O(1) space complexity, but the recursive approach uses the call stack. In the worst case (skewed tree), this is O(n). The actual space complexity is O(n) for the recursion stack, or O(log n) for a balanced tree.
• The problem mentions a follow-up to solve it both recursively and iteratively. While your recursive solution is correct, you could also implement the iterative version using a queue (BFS) or stack (DFS) to demonstrate both approaches.
• Minor: Consider adding a brief explanation of why the recursion works (mirror property: left.left mirrors right.right, and left.right mirrors right.left).

VERDICT: PASS