t3

[spencerkrebs]

No description provided.

[super30admin]

Root-to-leaf Sum II (pathsum2.py)

Your solution is well-structured and follows a clear recursive DFS approach. Here are some strengths and areas for improvement:

Strengths:

1. You correctly use backtracking with path.pop() to avoid storing multiple copies of the same path, which optimizes space usage during traversal.
2. You append a copy of the path when a valid path is found (using list(path)), which is necessary because the list is mutable and would be modified by subsequent pops.
3. The code is readable and follows good practices for recursive tree traversal.

Areas for Improvement:

1. Time Complexity Analysis: Your comment states "O(h) time", but the time complexity is actually O(n) because you visit each node once. The height h is only relevant for the recursive stack space. Additionally, when considering the cost of copying the path for each leaf node (which takes O(h) per copy), the worst-case time complexity becomes O(n * h) for a balanced tree or O(n^2) for a skewed tree. You should update your comment to reflect this.
2. Initialization of result: You initialize self.result as an instance variable. While this works, it might be better to avoid instance variables when they are not necessary. Consider using a local variable and passing it through parameters or using a closure to make the function more self-contained and avoid state in the class.
3. Default Parameters: Alternatively, you could define the helper function without making it a class method and use default parameters or nested functions to keep the result list local.

Suggested Improvement:
You can restructure the code to avoid using an instance variable for result. For example:

class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        result = []
        self.helper(root, targetSum, 0, [], result)
        return result
    
    def helper(self, root, targetSum, currSum, path, result):
        if root is None:
            return
        
        currSum += root.val
        path.append(root.val)

        if root.left is None and root.right is None:
            if currSum == targetSum:
                result.append(path[:])  # Append a copy
                
        self.helper(root.left, targetSum, currSum, path, result)
        self.helper(root.right, targetSum, currSum, path, result)
        path.pop()

Alternatively, you can use a nested function to avoid passing result as a parameter:

class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        result = []
        
        def helper(root, currSum, path):
            if root is None:
                return
            currSum += root.val
            path.append(root.val)
            if not root.left and not root.right and currSum == targetSum:
                result.append(path[:])
            helper(root.left, currSum, path)
            helper(root.right, currSum, path)
            path.pop()
            
        helper(root, 0, [])
        return result

This approach makes the code cleaner and avoids using instance variables for state that is only needed during the recursion.

Overall, your solution is correct and efficient, but the time complexity comment needs correction.

VERDICT: PASS

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Mirror Image of Itself Tree (symmetric-tree.py)

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VERDICT: NEEDS_IMPROVEMENT