Adaptive Matrix Traversal — Problem Statement & Solution Guide

TreesHardDepth-First Search

TimeO(N)

SpaceO(H)

Problem Description

You are given a binary tree represented by its root node. Each node contains an integer value. The goal is to compute the 'Adaptive Matrix Traversal' score, which is defined as the maximum sum of values along any path from the root to a leaf node. A path is considered valid only if it starts at the root and ends at a leaf (a node with no children). If the tree is empty, the score is 0.

The traversal must be adaptive in the sense that at each branching point, the algorithm must implicitly choose the branch that leads to the higher cumulative sum, effectively pruning the suboptimal paths without explicitly storing them. This requires a depth-first search strategy to explore all possible root-to-leaf paths and retain the maximum sum encountered.

Input: The root of a binary tree where each node has an integer value and optional left and right children.

Output: An integer representing the maximum sum of values from the root to any leaf node.

DSA Pattern Breakdown

"Adaptive Matrix Traversal"

hard

WHY DOES IT MATTER?

Root‑to‑leaf maximum sum is a fundamental pattern for tree DP and greedy decisions, teaching candidates how to combine local optimal choices into a global optimum while respecting hierarchical constraints.

OPTIMIZATION CHALLENGE

The key insight is recognizing that you never need to store all paths; you only need the best sum from each child, turning an exponential enumeration into a linear pass by exploiting the optimal substructure.

REAL-WORLD CONNECTION

In distributed systems, this mirrors finding the most profitable execution path through a workflow DAG where each node adds cost or revenue, and you must choose the path that maximizes net gain from start to termination.

During an interview, write a clean recursive helper that returns the max sum for a node; handle null nodes by returning 0 (or negative infinity if you need to differentiate empty subtrees) and remember to treat leaf nodes correctly to avoid adding extra zeros.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(H)

Core Theory — Why This Approach?

The Adaptive Matrix Traversal problem reduces to finding the maximum sum along any root‑to‑leaf path in a binary tree. This is a classic example of a tree DP (dynamic programming) problem where the optimal substructure property holds: the best path from a node to a leaf is the node's value plus the maximum of the best paths of its left and right subtrees. A depth‑first traversal (pre‑order, in‑order, or post‑order) can compute this recursively, propagating the best sum upward. Naïve approaches that enumerate all root‑to‑leaf paths quickly become infeasible because a balanced tree with height h has up to 2^h leaves, leading to exponential time. By visiting each node exactly once and using the return value of recursive calls, we achieve linear time. The optimal paradigm is a simple recursive DFS with backtracking, optionally converted to an iterative stack to control space usage, which respects the O(N) time and O(H) space where H is the tree height.

Interview Questions on This Problem

Q1How would you modify the algorithm to also return the actual path (list of node values) that yields the maximum root‑to‑leaf sum?

During the DFS, alongside the sum, return the path list from the current node to the leaf that gives the maximum sum. At each node, compare the sums from left and right children, pick the larger, prepend the current node's value to that child's path, and propagate upward. This adds O(H) extra space for the path but keeps overall complexity O(N).

Q2What changes are needed if the tree can contain negative values and you must treat a leaf as any node with no children, but you may also stop early if extending the path would decrease the sum?

The algorithm already handles negative values because it always chooses the maximum of left and right sub‑sums; however, to allow early termination, you can treat a node as a leaf if both child contributions are negative, returning the node's value alone. This ensures you never add a detrimental subtree, still preserving O(N) time.

Q3Explain how you would parallelize the computation of the maximum root‑to‑leaf sum on a massive distributed tree stored across multiple machines.

Partition the tree by subtrees rooted at a certain depth, assign each partition to a worker that computes the maximum sum for its subtree using the same DFS logic, then aggregate results at the parent level by adding the parent node's value to the maximum of its children's returned sums. This map‑reduce style reduces depth‑wise work and keeps communication proportional to the tree height.

Examples

Example 1

Input

root = [10, 5, 15, null, 7, 8, 20]

Output

Explanation: The tree has three root-to-leaf paths: 10->5->7 (sum=22), 10->15->8 (sum=33), and 10->15->20 (sum=45). Wait, let's re-verify the structure. If root=10, left=5, right=15. Left child of 5 is null, right child of 5 is 7. Left child of 15 is 8, right child of 15 is 20. Paths: 10->5->7 (sum=22), 10->15->8 (sum=33), 10->15->20 (sum=45). The maximum is 45. Let me correct the example to ensure clarity. Let's use a simpler tree. Root=1, Left=2, Right=3. Left of 2 is 4, Right of 2 is 5. Left of 3 is 6, Right of 3 is 7. Paths: 1->2->4 (7), 1->2->5 (8), 1->3->6 (10), 1->3->7 (11). Max is 11. Let's use this one. Corrected Example 1: Input: root = [1, 2, 3, 4, 5, 6, 7] Output: 11 Explanation: The root-to-leaf paths are: 1->2->4 (sum=7), 1->2->5 (sum=8), 1->3->6 (sum=10), 1->3->7 (sum=11). The maximum sum is 11.

Example 2

Input

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

Output

Explanation: The tree structure is: Root=5. Left child=-1, Right child=3. Left child of -1 is -2, Right child of -1 is -3. Left child of 3 is 4, Right child of 3 is 2. Paths: 5->-1->-2 (sum=2), 5->-1->-3 (sum=1), 5->3->4 (sum=12), 5->3->2 (sum=10). The maximum sum is 12. Wait, 5+3+4=12. Let me re-calculate. 5+3=8, 8+4=12. 5+3+2=10. 5-1-2=2. 5-1-3=1. Max is 12. Let's adjust the output to 12. Corrected Example 2: Input: root = [5, -1, 3, -2, -3, 4, 2] Output: 12 Explanation: The root-to-leaf paths are: 5->-1->-2 (sum=2), 5->-1->-3 (sum=1), 5->3->4 (sum=12), 5->3->2 (sum=10). The maximum sum is 12.

Example 3

Input

root = [100]

Output

Explanation: The tree consists of a single node which is both the root and a leaf. The only path is 100. The sum is 100.

Constraints

The number of nodes in the tree is between 1 and 10^5.
-10^9 <= Node.val <= 10^9.
The tree is a valid binary tree.
The depth of the tree is at most 10^5.

Optimal Approach & Strategy

Perform a single DFS, returning the maximum sum from each node to a leaf; combine child results with the node's value to propagate the best sum upward, achieving linear time.

Brute Force Approach

Enumerate every root‑to‑leaf path, compute its sum, and keep the maximum; this requires exploring all paths which is exponential in the tree height.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

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

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

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

def solution(nums):
   n = len(nums)
   dp = [[0]*n for _ in range(n)]
   max_sum = float('-inf')
   for i in range(n):
       for j in range(i, n):
           sum_val = 0
           for k in range(i, j+1):
               sum_val += nums[k]
               dp[i][j] = max(dp[i][j], sum_val)
               max_sum = max(max_sum, dp[i][j])
   return max_sum

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

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Adaptive Matrix Traversal 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 Matrix Traversal — 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