Completed Leetcode 113 and 101

[allurkarsneha]

No description provided.

[super30admin]

Path Sum II (Problem 1- leetcode113.py)

Strengths:

1. Excellent use of backtracking with path.pop() - this is more memory-efficient than the reference solution's approach of passing a copy of the path
2. Clean, readable code with good structure
3. Correct handling of all edge cases
4. Proper use of list(path) to create a copy when adding to results

Areas for Improvement:

1. Complexity Analysis: Your time and space complexity comments are inaccurate. The actual time complexity is O(n²) in the worst case (skewed tree) because copying each path takes O(n) time. The space complexity is also O(n²) when considering the output storage. Update your comments to reflect this.
2. Minor optimization: You could add an early termination check - if currSum exceeds targetSum and all values are positive, you could prune. However, since values can be negative (as per constraints), this optimization isn't always valid, so your current approach is fine.
3. Documentation: Consider adding a brief docstring to the helper method explaining its purpose.

VERDICT: PASS

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Symmetric Tree (Problem 2- Leetcode101.py)

The student's solution uses a recursive approach to check if a binary tree is symmetric. Let me analyze it:

1. Correctness: The solution correctly handles all cases:

   • Both nodes are None → returns True
   • One node is None → returns False
   • Values don't match → returns False
   • Recursive calls check left.left vs right.right and left.right vs right.left

   However, there's a potential issue: if root is None, the code will crash with AttributeError because root.left is accessed without checking if root is None. The reference solution handles this with if(root == nullptr) return true;. While the problem constraints state the number of nodes is in range [1, 1000], so root won't be None, it's still a good practice to handle this edge case.

2. Time Complexity: O(n) - same as the reference solution. Each node is visited once.

3. Space Complexity: O(h) where h is the height of the tree (due to recursion stack). This is actually better than the reference solution's O(n) in the best case (balanced tree), but could be worse in the worst case (skewed tree) where h = n.

4. Code Quality:

   • Well-structured with a separate helper function
   • Clear variable naming
   • Good comments on time/space complexity
   • Follows Python best practices
   • The recursive approach is clean and readable

5. Efficiency: The solution is efficient. The recursive approach is a classic and elegant way to solve this problem.

The solution is essentially correct and solves the problem properly. The only minor concern is the missing edge case for when root is None, but given the problem constraints, this isn't a critical issue.

FEEDоBACK:
Strengths:

• Clean, recursive implementation that correctly solves the problem
• Good separation of concerns with the helper function
• Proper handling of all the key cases (both None, one None, value mismatch)
• Well-documented with time and space complexity comments
• The recursive approach is elegant and easy to understand

Areas for improvement:

• Consider adding a base case for when root is None at the start of isSymmetric, even though the constraints suggest it won't happen. This makes the function more robust:

  def isSymmetric(self, root):
      if root is None:
          return True
      return self.helper(root.left, root.right)

• The problem mentions a follow-up to solve it both recursively and iteratively. While you provided the recursive solution, you could also implement the iterative version using a queue/stack for practice.

Overall, this is a solid solution that demonstrates a good understanding of the problem and recursive thinking.

VERDICT: PASS