Adaptive Subsequence Sum — Problem Statement & Solution Guide

TreesEasyDepth-First Search

TimeO(N)

SpaceO(H)

Problem Description

You are given a binary tree where each node contains an integer value. The 'Adaptive Subsequence Sum' of a tree is defined as the sum of values along any root-to-leaf path, but with an adaptive rule: if a node's value is negative, it is excluded from the sum for that specific path calculation, effectively treating it as zero for the purpose of accumulation. However, the path must still traverse through the node to reach its children. Your task is to compute the maximum possible Adaptive Subsequence Sum among all root-to-leaf paths in the tree.

A leaf node is defined as a node that has no children. If the tree is empty, the sum is 0. The adaptive rule applies independently to each node encountered during the traversal: positive or zero values are added to the running sum, while negative values are skipped (added as 0) but do not terminate the path.

Input: The root of a binary tree.

Output: An integer representing the maximum adaptive sum from the root to any leaf.

DSA Pattern Breakdown

"Adaptive Subsequence Sum"

easy

WHY DOES IT MATTER?

The pattern exemplifies tree traversal with stateful accumulation, a core technique for many hierarchical data problems such as file‑system quotas, organizational budgets, and decision‑tree evaluations where certain nodes may be ignored or weighted differently.

OPTIMIZATION CHALLENGE

The key insight is recognizing that negative values can be clamped to zero locally, eliminating the need to explore alternative subsets of nodes. This reduces the exponential path enumeration to a single linear pass.

REAL-WORLD CONNECTION

Consider a network of microservices where each service contributes latency; negative latency (e.g., caching) is treated as zero for end‑to‑end SLA calculations. Computing the worst‑case request path mirrors the Adaptive Subsequence Sum traversal.

During an interview, write the DFS helper that returns the maximum adaptive sum from the current node to any leaf; this isolates the recursion and makes it easy to reason about base cases and the clipping operation.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(H)

Core Theory — Why This Approach?

The Adaptive Subsequence Sum problem is a variant of the classic root‑to‑leaf path sum on a binary tree. The twist is that any node with a negative value is treated as zero for the purpose of the accumulation, though the traversal must still include that node. A naive solution would enumerate every root‑to‑leaf path, compute the adjusted sum for each, and keep the maximum. This approach is O(2^h) in the worst case because a balanced tree of height h has roughly 2^h leaves, leading to exponential time on large inputs.

The optimal paradigm leverages depth‑first search (DFS) with a simple state‑carrying variable: the current adaptive sum. As we recurse, we add max(0, node.val) to the running total, ensuring negatives contribute nothing. When a leaf is reached, we compare the accumulated sum with a global maximum. This yields a linear O(N) traversal where N is the number of nodes, because each node is visited exactly once. The space usage is O(H) for the recursion stack, where H is the tree height, which in the worst case (skewed tree) equals O(N) but is optimal for a DFS solution.

Why this works efficiently is that the problem exhibits optimal substructure: the best adaptive sum for any subtree depends only on the best sums of its children and the current node’s contribution (clipped at zero). Thus, a single pass suffices, and memoization is unnecessary. This mirrors many tree DP problems where local decisions propagate upward without recomputation.

Interview Questions on This Problem

Q1How would you modify the solution if the adaptive rule required treating any node with value less than a given threshold K as zero?

Replace max(0, node.val) with max(K, node.val) during the DFS accumulation. The rest of the algorithm remains unchanged, still running in O(N) time and O(H) space.

Q2Can you compute the Adaptive Subsequence Sum iteratively without recursion? What data structure would you use?

Yes, perform an explicit stack‑based DFS or a BFS using a queue, storing pairs of (node, currentAdaptiveSum). For each popped node, push its children with updated sums. This avoids recursion depth limits while preserving O(N) time and O(H) auxiliary space.

Q3If the tree is extremely large and stored in a distributed key‑value store, how would you design a MapReduce job to compute the maximum adaptive root‑to‑leaf sum?

First, emit each node with its parent identifier. In the map phase, propagate partial sums from root downwards, clipping negatives to zero. The reduce phase aggregates child contributions, keeping the maximum sum reaching each leaf. Finally, a second reduce step extracts the global maximum. This mimics a parallel DFS across partitions, still O(N) overall work.

Examples

Example 1

Input

root = [5, -3, 2, null, 4, null, 1]

Output

Explanation: Path 1: 5 -> -3 -> 4. Sum = 5 + 0 + 4 = 9. Path 2: 5 -> 2 -> 1. Sum = 5 + 2 + 1 = 8. Wait, let's re-evaluate. Path 1: 5 (add 5), -3 (add 0), 4 (add 4) => 9. Path 2: 5 (add 5), 2 (add 2), 1 (add 1) => 8. Max is 9. Let me correct the example to be clearer. Let's use a different tree. Revised Example 1: Input: root = [10, -5, 3, null, 2, null, 4] Path 1: 10 -> -5 -> 2. Sum = 10 + 0 + 2 = 12. Path 2: 10 -> 3 -> 4. Sum = 10 + 3 + 4 = 17. Max is 17. Let's stick to the first one but fix the explanation. Input: root = [5, -3, 2, null, 4, null, 1] Path 1: 5 -> -3 -> 4. Sum = 5 + 0 + 4 = 9. Path 2: 5 -> 2 -> 1. Sum = 5 + 2 + 1 = 8. Output: 9.

Example 2

Input

root = [-1, -2, -3]

Output

Explanation: Path 1: -1 -> -2. Sum = 0 + 0 = 0. Path 2: -1 -> -3. Sum = 0 + 0 = 0. The maximum sum is 0.

Example 3

Input

root = [1, 2, 3, 4, 5, 6, 7]

Output

Explanation: All values are positive, so the adaptive rule has no effect. Path 1: 1->2->4 = 7. Path 2: 1->2->5 = 8. Path 3: 1->3->6 = 10. Path 4: 1->3->7 = 11. Wait, 1+3+7=11. Let's check 1+2+5=8. 1+3+6=10. Max is 11. Let me re-calculate. 1+3+7 = 11. 1+2+5 = 8. 1+2+4 = 7. 1+3+6 = 10. Max is 11. I will adjust the output to 11.

Constraints

The number of nodes in the tree is in the range [1, 10^5].
-10^9 <= Node.val <= 10^9
The tree is a valid binary tree.
The depth of the tree will not exceed 10^5.

Optimal Approach & Strategy

Perform a single DFS, accumulating max(0, node.val) along the way and updating a global maximum at each leaf; this runs in linear time.

Brute Force Approach

Enumerate every root‑to‑leaf path, compute the adjusted sum by skipping negatives, and keep the maximum; this is exponential in the height of the tree.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

function solution(nums) {
   let maxSum = 0;
   let dp = new Array(nums.length).fill(0);
   for (let i = 0; i < nums.length; i++) {
       let sum = 0;
       for (let j = i; j < nums.length; j++) {
           sum += nums[j];
           dp[j] = Math.max(dp[j], sum);
           maxSum = Math.max(maxSum, dp[j]);
       }
   }
   return maxSum;
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       int maxSum = 0;
       vector<int> dp(nums.size());
       for (int i = 0; i < nums.size(); i++) {
           int sum = 0;
           for (int j = i; j < nums.size(); j++) {
               sum += nums[j];
               dp[j] = max(dp[j], sum);
               maxSum = max(maxSum, dp[j]);
           }
       }
       return maxSum;
   }
};

class Solution {
   public int solution(int[] nums) {
       int maxSum = 0;
       int[] dp = new int[nums.length];
       for (int i = 0; i < nums.length; i++) {
           int sum = 0;
           for (int j = i; j < nums.length; j++) {
               sum += nums[j];
               dp[j] = Math.max(dp[j], sum);
               maxSum = Math.max(maxSum, dp[j]);
           }
       }
       return maxSum;
   }
}

def solution(nums):
   max_sum = 0
   dp = [0] * len(nums)
   for i in range(len(nums)):
       sum_val = 0
       for j in range(i, len(nums)):
           sum_val += nums[j]
           dp[j] = max(dp[j], sum_val)
           max_sum = max(max_sum, dp[j])
   return max_sum

function solution(nums) {
   let maxSum = 0;
   let dp = new Array(nums.length).fill(0);
   for (let i = 0; i < nums.length; i++) {
       let sum = 0;
       for (let j = i; j < nums.length; j++) {
           sum += nums[j];
           dp[j] = Math.max(dp[j], sum);
           maxSum = Math.max(maxSum, dp[j]);
       }
   }
   return maxSum;
}

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Adaptive Subsequence Sum Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Trees

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Adaptive Subsequence Sum — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor