Is Path Sum III easy or hard?

Path Sum III is generally classified as a medium-level problem. The brute-force DFS idea is easy to understand, but recognizing the prefix sum optimization and managing the hash map correctly during recursion requires stronger problem-solving experience.

How to solve Path Sum III in O(n)?

Traverse the tree using DFS while maintaining a running prefix sum from the root to the current node. Store counts of prefix sums in a hash map. At each node, check how many times prefixSum - targetSum has appeared, which represents valid paths ending at that node. Update the map during traversal and remove entries during backtracking.

What is the best approach for Path Sum III?

The most efficient approach uses a prefix sum hash map during a depth-first traversal. While exploring the tree, track the running sum and check if a previous prefix sum equal to currentSum - target exists. This allows you to count valid paths in O(n) time because each node is processed once.

Is Path Sum III asked at Google/Amazon/Meta?

Path Sum III is a common binary tree interview problem and has appeared in interviews at companies like Amazon, Meta, and Google. It tests understanding of DFS traversal, prefix sums, and how to maintain state across recursive tree paths.

What data structure is used in Path Sum III?

The optimal solution uses a hash map to store prefix sums and their frequencies during a depth-first traversal of the binary tree. The tree itself is explored using recursion or a DFS stack, while the hash map enables constant-time prefix sum lookups.

What is the time complexity of Path Sum III?

The brute-force DFS approach runs in O(n^2) time in the worst case because a DFS is started from every node. The optimized prefix sum method reduces this to O(n) by storing previously seen running sums in a hash map and performing constant-time lookups.

Path Sum III Python or Java solution approach?

Both Python and Java implementations follow the same pattern: perform DFS while tracking the current prefix sum and store counts in a hash map (dictionary in Python, HashMap in Java). Each node checks if prefixSum - target exists in the map to count valid paths.

Description Solution Video 15:02

437. Path Sum III

Medium46.6% acceptance11 companies

Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.

The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).

Example 1:

Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
Output: 3
Explanation: The paths that sum to 8 are shown.

Example 2:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: 3

Constraints:

The number of nodes in the tree is in the range [0, 1000].
-10⁹ <= Node.val <= 10⁹
-1000 <= targetSum <= 1000

Asked at 11 companies

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437. Path Sum III

Asked at 11 companies

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