Path Sum III (Path in Tree)
Given the root of a binary tree and an integer `targetSum`, return the number of paths where the sum of the values equals `targetSum`. The path does not need to start or end at the root or a leaf. Use a HashMap for efficient path tracking.
Why Interviewers Ask This
Google interviewers ask this question to evaluate a candidate's ability to optimize brute-force tree traversals using advanced data structures. They specifically test your understanding of prefix sums and hash map usage to reduce time complexity from O(N^2) to O(N). It assesses whether you can handle non-standard path constraints where paths start and end anywhere, requiring deep insight into recursive state management.
How to Answer This Question
1. Clarify the problem scope immediately: confirm that paths can start at any node and end at any descendant, not just root-to-leaf.
2. Propose a brute-force solution first to establish a baseline, explaining its O(N^2) complexity, then pivot to the optimized approach.
3. Introduce the Prefix Sum technique with a HashMap to store cumulative sums encountered along the current recursion stack.
4. Detail the algorithm: calculate current prefix sum, check if (currentSum - targetSum) exists in the map, increment count if found, then recursively process children.
5. Emphasize the critical backtracking step: remove the current prefix sum from the map upon returning from recursion to ensure the map only contains valid ancestors for subsequent branches.
Key Points to Cover
- Explicitly mention reducing time complexity from O(N^2) to O(N) using prefix sums
- Demonstrate clear understanding of backtracking to maintain correct map state across different branches
- Explain how the HashMap stores frequencies of prefix sums rather than just existence
- Clarify that the path must be downward, connecting a node to one of its descendants
- Show awareness of edge cases like empty trees or negative numbers affecting the sum logic
Sample Answer
To solve the Path Sum III problem efficiently, I would first acknowledge that a naive approach checking every node as a potential start point would result in O(N^2) time complexity. Since Google values scalable solutions, I aim for an O(N) approach using a HashMap to track prefix sums.
My strategy involves a single depth-first traversal while maintaining a running sum from the root to the current node. As we visit each node, we calculate the current cumulative sum. We then check if there exists a previous prefix sum such that (currentSum - previousSum) equals our target. If (currentSum - targetSum) exists in our HashMap, it means a valid path ending at the current node has been found, so we add the frequency of that difference to our total count.
Crucially, before recursing into child nodes, we update the HashMap with the current sum. After processing both left and right subtrees, we must backtrack by decrementing the count of the current sum in the map or removing it entirely. This ensures the map accurately reflects only the path from the root to the current parent, preventing invalid cross-branch path calculations. This method allows us to find all valid paths in linear time, handling the constraint that paths need not start at the root or end at a leaf.
Common Mistakes to Avoid
- Failing to remove the current prefix sum from the HashMap after recursion, causing incorrect counts in sibling branches
- Assuming paths must start at the root or end at a leaf, ignoring the specific problem constraints
- Using a simple set instead of a HashMap, which fails when multiple paths share the same prefix sum value
- Not handling the case where the current node's value itself equals the target sum without a preceding prefix
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