LeetCode #2925 — MEDIUM

Maximum Score After Applying Operations on a Tree

Move from brute-force thinking to an efficient approach using dynamic programming strategy.

The Problem

Problem Statement

There is an undirected tree with n nodes labeled from 0 to n - 1, and rooted at node 0. You are given a 2D integer array edges of length n - 1, where edges[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree.

You are also given a 0-indexed integer array values of length n, where values[i] is the value associated with the i^th node.

You start with a score of 0. In one operation, you can:

Pick any node i.
Add values[i] to your score.
Set values[i] to 0.

A tree is healthy if the sum of values on the path from the root to any leaf node is different than zero.

Return the maximum score you can obtain after performing these operations on the tree any number of times so that it remains healthy.

Example 1:

Input: edges = [[0,1],[0,2],[0,3],[2,4],[4,5]], values = [5,2,5,2,1,1]
Output: 11
Explanation: We can choose nodes 1, 2, 3, 4, and 5. The value of the root is non-zero. Hence, the sum of values on the path from the root to any leaf is different than zero. Therefore, the tree is healthy and the score is values[1] + values[2] + values[3] + values[4] + values[5] = 11.
It can be shown that 11 is the maximum score obtainable after any number of operations on the tree.

Example 2:

Input: edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]], values = [20,10,9,7,4,3,5]
Output: 40
Explanation: We can choose nodes 0, 2, 3, and 4.
- The sum of values on the path from 0 to 4 is equal to 10.
- The sum of values on the path from 0 to 3 is equal to 10.
- The sum of values on the path from 0 to 5 is equal to 3.
- The sum of values on the path from 0 to 6 is equal to 5.
Therefore, the tree is healthy and the score is values[0] + values[2] + values[3] + values[4] = 40.
It can be shown that 40 is the maximum score obtainable after any number of operations on the tree.

Constraints:

2 <= n <= 2 * 10⁴
edges.length == n - 1
edges[i].length == 2
0 <= a_i, b_i < n
values.length == n
1 <= values[i] <= 10⁹
The input is generated such that edges represents a valid tree.

Step 01

Brute Force Baseline

Problem summary: There is an undirected tree with n nodes labeled from 0 to n - 1, and rooted at node 0. You are given a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. You are also given a 0-indexed integer array values of length n, where values[i] is the value associated with the ith node. You start with a score of 0. In one operation, you can: Pick any node i. Add values[i] to your score. Set values[i] to 0. A tree is healthy if the sum of values on the path from the root to any leaf node is different than zero. Return the maximum score you can obtain after performing these operations on the tree any number of times so that it remains healthy.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming · Tree

Example 1

[[0,1],[0,2],[0,3],[2,4],[4,5]]
[5,2,5,2,1,1]

Example 2

[[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]]
[20,10,9,7,4,3,5]

Core Insight

What unlocks the optimal approach

Let <code>dp[i]</code> be the maximum score we can get on the subtree rooted at <code>i</code> and <code>sum[i]</code> be the sum of all the values of the subtree rooted at <code>i</code>.
If we don’t take <code>value[i]</code> into the final score, we can take all the nodes of the subtrees rooted at <code>i</code>’s children.
If we take <code>value[i]</code> into the score, then each subtree rooted at its children should satisfy the constraints.
<code>dp[x] = max(value[x] + sigma(dp[y]), sigma(sum[y]))</code>, where <code>y</code> is a direct child of <code>x</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2925: Maximum Score After Applying Operations on a Tree
class Solution {
    private List<Integer>[] g;
    private int[] values;

    public long maximumScoreAfterOperations(int[][] edges, int[] values) {
        int n = values.length;
        g = new List[n];
        this.values = values;
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        return dfs(0, -1)[1];
    }

    private long[] dfs(int i, int fa) {
        long a = 0, b = 0;
        boolean leaf = true;
        for (int j : g[i]) {
            if (j != fa) {
                leaf = false;
                var t = dfs(j, i);
                a += t[0];
                b += t[1];
            }
        }
        if (leaf) {
            return new long[] {values[i], 0};
        }
        return new long[] {values[i] + a, Math.max(values[i] + b, a)};
    }
}

// Accepted solution for LeetCode #2925: Maximum Score After Applying Operations on a Tree
func maximumScoreAfterOperations(edges [][]int, values []int) int64 {
	g := make([][]int, len(values))
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	var dfs func(int, int) (int64, int64)
	dfs = func(i, fa int) (int64, int64) {
		a, b := int64(0), int64(0)
		leaf := true
		for _, j := range g[i] {
			if j != fa {
				leaf = false
				aa, bb := dfs(j, i)
				a += aa
				b += bb
			}
		}
		if leaf {
			return int64(values[i]), int64(0)
		}
		return int64(values[i]) + a, max(int64(values[i])+b, a)
	}
	_, b := dfs(0, -1)
	return b
}

# Accepted solution for LeetCode #2925: Maximum Score After Applying Operations on a Tree
class Solution:
    def maximumScoreAfterOperations(
        self, edges: List[List[int]], values: List[int]
    ) -> int:
        def dfs(i: int, fa: int = -1) -> (int, int):
            a = b = 0
            leaf = True
            for j in g[i]:
                if j != fa:
                    leaf = False
                    aa, bb = dfs(j, i)
                    a += aa
                    b += bb
            if leaf:
                return values[i], 0
            return values[i] + a, max(values[i] + b, a)

        g = [[] for _ in range(len(values))]
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        return dfs(0)[1]

// Accepted solution for LeetCode #2925: Maximum Score After Applying Operations on a Tree
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2925: Maximum Score After Applying Operations on a Tree
// class Solution {
//     private List<Integer>[] g;
//     private int[] values;
// 
//     public long maximumScoreAfterOperations(int[][] edges, int[] values) {
//         int n = values.length;
//         g = new List[n];
//         this.values = values;
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (var e : edges) {
//             int a = e[0], b = e[1];
//             g[a].add(b);
//             g[b].add(a);
//         }
//         return dfs(0, -1)[1];
//     }
// 
//     private long[] dfs(int i, int fa) {
//         long a = 0, b = 0;
//         boolean leaf = true;
//         for (int j : g[i]) {
//             if (j != fa) {
//                 leaf = false;
//                 var t = dfs(j, i);
//                 a += t[0];
//                 b += t[1];
//             }
//         }
//         if (leaf) {
//             return new long[] {values[i], 0};
//         }
//         return new long[] {values[i] + a, Math.max(values[i] + b, a)};
//     }
// }

// Accepted solution for LeetCode #2925: Maximum Score After Applying Operations on a Tree
function maximumScoreAfterOperations(edges: number[][], values: number[]): number {
    const g: number[][] = Array.from({ length: values.length }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    const dfs = (i: number, fa: number): [number, number] => {
        let [a, b] = [0, 0];
        let leaf = true;
        for (const j of g[i]) {
            if (j !== fa) {
                const [aa, bb] = dfs(j, i);
                a += aa;
                b += bb;
                leaf = false;
            }
        }
        if (leaf) {
            return [values[i], 0];
        }
        return [values[i] + a, Math.max(values[i] + b, a)];
    };
    return dfs(0, -1)[1];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.