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Maximum Score After Applying Operations on a Tree solution and explanation

MediumDepth-First SearchDynamic ProgrammingTree8 min read47.2% acceptanceUpdated 9 Apr 2026
Asked atGoogle
Practise Maximum Score After Applying Operations on a Tree
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Leetcode weekly contest 370 solution | Maximum Score After Applying Operations on a Tree | HindiPawan Kumar Giri · 27:30 · 1K viewsWatch on YouTube
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The problem

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.

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 <= ai, bi < n
  • values.length == n
  • 1 <= values[i] <= 109
  • The input is generated such that edges represents a valid tree.

Approach overview

Problem Overview: You are given a tree where each node has a value. You may apply operations that collect a node’s value into the score and set that node to zero. After all operations, every root-to-leaf path must still have a non-zero total. The goal is to maximize the score collected.

Approach 1: DFS with Tree Dynamic Programming (O(n) time, O(n) space)

The key observation: instead of directly maximizing the score, compute the minimum value that must remain in the tree so every root-to-leaf path keeps a positive sum. If you know the total sum of all node values, the final score is simply totalSum - requiredSum. Use Depth-First Search to process the tree bottom‑up. For a leaf node, you must keep its value because removing it would make the path sum zero. For an internal node, you have a choice: keep the node’s value, or rely on its children to keep enough value in the subtree. Therefore the minimum required value at a node is min(nodeValue, sum(childRequired)). This recursive decision naturally forms a Dynamic Programming state on the tree. Traverse the tree, compute the required value for each subtree, and subtract the root’s required value from the total sum.

Approach 2: Iterative DFS Using Stack (O(n) time, O(n) space)

The same DP logic can be implemented iteratively to avoid recursion depth limits. Build an adjacency list and simulate post‑order traversal with a stack. Each stack frame tracks whether children have been processed. Once all children of a node are evaluated, compute its required value using the same rule: leaf nodes return their value, while internal nodes return min(nodeValue, sum(childRequired)). Store intermediate results in an array or map indexed by node. This approach is useful in environments where recursion depth may exceed limits or when you prefer explicit control over traversal.

Recommended for interviews: The recursive DFS with tree DP is the expected solution. It runs in linear time, clearly models the constraint that each path must retain value, and demonstrates comfort with tree-based dynamic programming. Mentioning the iterative DFS variant shows awareness of stack-based traversal and recursion limits.

Approach 1: DFS and Tree Traversal

Use Depth First Search (DFS) to traverse the tree and consider each node's contribution to potential scores while ensuring the tree remains healthy. The DFS approach allows us to explore all nodes by starting at the root and visiting child nodes before visiting sibling nodes, keeping track of the path value and the chosen nodes.

def dfs(node, current_path_sum, visited, adj, values):
    visited[node] = True
    current_path_sum += values[node]

    global max_score
    isLeaf = True

    for neighbor in adj[node]:
        if not visited[neighbor]:
            isLeaf = False
            dfs(neighbor, current_path_sum, visited, adj, values)

    if isLeaf and current_path_sum != 0:
        max_score = max(max_score, current_path_sum)

    visited[node] = False


n = 6
edges = [[0, 1], [0, 2], [0, 3], [2, 4], [4, 5]]
values = [5, 2, 5, 2, 1, 1]
adj = [[] for _ in range(n)]
for a, b in edges:
    adj[a].append(b)
    adj[b].append(a)
visited = [False] * n

max_score = 0

# Start DFS from node 0 with initial path sum of 0
dfs(0, 0, visited, adj, values)

print(f'Maximum Score: {max_score}')
import java.util.*;

public class Main {
    static List<Integer>[] adj;
    static int[] values;
    static boolean[] visited;
    static int maxScore;

    public static void dfs(int node, int currentPathSum) {
        visited[node] = true;
        currentPathSum += values[node];

        boolean isLeaf = true;
        for (int neighbor : adj[node]) {
            if (!visited[neighbor]) {
                isLeaf = false;
                dfs(neighbor, currentPathSum);
            }
        }

        if (isLeaf && currentPathSum != 0) {
            maxScore = Math.max(maxScore, currentPathSum);
        }
        visited[node] = false;
    }

    public static void main(String[] args) {
        int n = 6;
        int[][] edges = {{0, 1}, {0, 2}, {0, 3}, {2, 4}, {4, 5}};
        values = new int[]{5, 2, 5, 2, 1, 1};
        adj = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<>();
        }
        visited = new boolean[n];

        for (int[] edge : edges) {
            adj[edge[0]].add(edge[1]);
            adj[edge[1]].add(edge[0]);
        }

        maxScore = 0;
        dfs(0, 0);

        System.out.println("Maximum Score: " + maxScore);
    }
}
#include <iostream>
#include <vector>
using namespace std;

vector<vector<int>> adj;
vector<int> values;
vector<bool> visited;
int max_score = 0;

void dfs(int node, int current_path_sum) {
    visited[node] = true;
    current_path_sum += values[node];

    bool isLeaf = true;
    for (int neighbor : adj[node]) {
        if (!visited[neighbor]) {
            isLeaf = false;
            dfs(neighbor, current_path_sum);
        }
    }
    
    if (isLeaf && current_path_sum != 0) {
        max_score = max(max_score, current_path_sum);
    }
    visited[node] = false;
}

int main() {
    // Example input
    int n = 6;
    vector<pair<int, int>> edges = {{0, 1}, {0, 2}, {0, 3}, {2, 4}, {4, 5}};
    values = {5, 2, 5, 2, 1, 1};
    adj.resize(n);
    visited.resize(n, false);

    for (auto edge : edges) {
        adj[edge.first].push_back(edge.second);
        adj[edge.second].push_back(edge.first);
    }

    dfs(0, 0);

    cout << "Maximum Score: " << max_score << endl;
    return 0;
}
const dfs = (node, currentPathSum, visited, adj, values) => {
    visited[node] = true;
    currentPathSum += values[node];

    let isLeaf = true;
    for (let neighbor of adj[node]) {
        if (!visited[neighbor]) {
            isLeaf = false;
            dfs(neighbor, currentPathSum, visited, adj, values);
        }
    }

    if (isLeaf && currentPathSum !== 0) {
        maxScore = Math.max(maxScore, currentPathSum);
    }
    visited[node] = false;
};

let n = 6;
let edges = [[0, 1], [0, 2], [0, 3], [2, 4], [4, 5]];
let values = [5, 2, 5, 2, 1, 1];
let adj = Array.from({ length: n }, () => []);
for (let [a, b] of edges) {
    adj[a].push(b);
    adj[b].push(a);
}
let visited = Array(n).fill(false);

let maxScore = 0;

dfs(0, 0, visited, adj, values);

console.log(`Maximum Score: ${maxScore}`);
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAXN 20000

int values[MAXN];
int adj[MAXN][MAXN];
int visited[MAXN];
int current_path_sum;
int max_score;

void dfs(int node, int n) {
    visited[node] = 1;
    current_path_sum += values[node];
    
    for (int i = 0; i < n; i++) {
        if (adj[node][i] && !visited[i]) {
            dfs(i, n);
        }
    }
    
    // Update max score and reset path if not valid
    if (current_path_sum != 0) {
        if (max_score < current_path_sum) {
            max_score = current_path_sum;
        }
        current_path_sum = 0;
    }
    visited[node] = 0;  // Unvisit for other paths
}

int main() {
    int n;
    // Example input
    int edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {2, 4}, {4, 5}};
    int values_in[] = {5, 2, 5, 2, 1, 1};
    n = 6;

    memcpy(values, values_in, sizeof(values_in));
    memset(visited, 0, sizeof(visited));
    memset(adj, 0, sizeof(adj));

    for (int i = 0; i < n - 1; i++) {
        adj[edges[i][0]][edges[i][1]] = 1;
        adj[edges[i][1]][edges[i][0]] = 1;
    }

    max_score = 0;
    for (int i = 0; i < n; i++) {
        current_path_sum = 0;
        dfs(i, n);
    }

    printf("Maximum Score: %d\n", max_score);
    return 0;
}
using System;
using System.Collections.Generic;

class TreeMaxScore {
    static List<int>[] adj;
    static int[] values;
    static bool[] visited;
    static int maxScore;

    static void DFS(int node, int currentPathSum) {
        visited[node] = true;
        currentPathSum += values[node];

        bool isLeaf = true;
        foreach (var neighbor in adj[node]) {
            if (!visited[neighbor]) {
                isLeaf = false;
                DFS(neighbor, currentPathSum);
            }
        }

        if (isLeaf && currentPathSum != 0) {
            maxScore = Math.Max(maxScore, currentPathSum);
        }
        visited[node] = false;
    }

    public static void Main() {
        int n = 6;
        int[][] edges = { new[]{0, 1}, new[]{0, 2}, new[]{0, 3}, new[]{2, 4}, new[]{4, 5} };
        values = new[] { 5, 2, 5, 2, 1, 1 };
        adj = new List<int>[n];
        visited = new bool[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new List<int>();
        }

        foreach (var edge in edges) {
            adj[edge[0]].Add(edge[1]);
            adj[edge[1]].Add(edge[0]);
        }

        maxScore = 0;
        DFS(0, 0);

        Console.WriteLine("Maximum Score: " + maxScore);
    }
}

The Python solution uses DFS to navigate through the tree structure, updating the path sum as each node is visited and recalculated. If a path yields a healthy score, it updates the maximum score found.

Complexity

Time Complexity: O(n) - The tree is explored entirely once.
Space Complexity: O(n) - Recursive stack used and adjacency list utilized.

Try this approach in the editor

Approach 2: Iterative DFS using Stack

An alternative to recursive DFS is using an iterative approach with a stack. This can help avoid recursion limitations in programming languages by simulating the call stack manually.

def dfs_iterative(adj, values, n):
    visited = [False] * n
    stack = [(0, values[0])]
    global max_score
    max_score = 0

    while stack:
        node, current_path_sum = stack.pop()
        visited[node] = True
        isLeaf = True

        for neighbor in adj[node]:
            if not visited[neighbor]:
                isLeaf = False
                stack.append((neighbor, current_path_sum + values[neighbor]))

        if isLeaf and current_path_sum != 0:
            max_score = max(max_score, current_path_sum)

n = 6
edges = [[0, 1], [0, 2], [0, 3], [2, 4], [4, 5]]
values = [5, 2, 5, 2, 1, 1]
adj = [[] for _ in range(n)]
for a, b in edges:
    adj[a].append(b)
    adj[b].append(a)

max_score = 0
dfs_iterative(adj, values, n)
print('Maximum Score:', max_score)
import java.util.*;

public class Main {
    static int maxScore;

    public static void dfsIterative(List<Integer>[] adj, int[] values, int n) {
        boolean[] visited = new boolean[n];
        Stack<int[]> stack = new Stack<>();
        stack.push(new int[]{0, values[0]});

        while (!stack.isEmpty()) {
            int[] curr = stack.pop();
            int node = curr[0];
            int currentPathSum = curr[1];
            visited[node] = true;

            boolean isLeaf = true;
            for (int neighbor : adj[node]) {
                if (!visited[neighbor]) {
                    isLeaf = false;
                    stack.push(new int[]{neighbor, currentPathSum + values[neighbor]});
                }
            }

            if (isLeaf && currentPathSum != 0) {
                maxScore = Math.max(maxScore, currentPathSum);
            }
        }
    }

    public static void main(String[] args) {
        int n = 6;
        int[][] edges = {{0, 1}, {0, 2}, {0, 3}, {2, 4}, {4, 5}};
        int[] values = {5, 2, 5, 2, 1, 1};
        List<Integer>[] adj = new List[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<>();
        }

        for (int[] edge : edges) {
            adj[edge[0]].add(edge[1]);
            adj[edge[1]].add(edge[0]);
        }

        maxScore = 0;
        dfsIterative(adj, values, n);

        System.out.println("Maximum Score: " + maxScore);
    }
}
#include <iostream>
#include <vector>
#include <stack>
using namespace std;

int max_score = 0;

void dfsIterative(vector<vector<int>>& adj, vector<int>& values, int n) {
    vector<bool> visited(n, false);
    stack<pair<int, int>> st;
    st.push({0, values[0]});
    
    while (!st.empty()) {
        auto [node, currentPathSum] = st.top();
        st.pop();
        visited[node] = true;

        bool isLeaf = true;
        for (int neighbor : adj[node]) {
            if (!visited[neighbor]) {
                isLeaf = false;
                st.push({neighbor, currentPathSum + values[neighbor]});
            }
        }

        if (isLeaf && currentPathSum != 0) {
            max_score = max(max_score, currentPathSum);
        }
    }
}

int main() {
    int n = 6;
    vector<pair<int, int>> edges = {{0, 1}, {0, 2}, {0, 3}, {2, 4}, {4, 5}};
    vector<int> values = {5, 2, 5, 2, 1, 1};
    vector<vector<int>> adj(n);

    for (auto edge : edges) {
        adj[edge.first].push_back(edge.second);
        adj[edge.second].push_back(edge.first);
    }

    dfsIterative(adj, values, n);

    cout << "Maximum Score: " << max_score << endl;
    return 0;
}
const dfsIterative = (adj, values, n) => {
    let visited = Array(n).fill(false);
    let stack = [[0, values[0]]];

    while (stack.length > 0) {
        let [node, currentPathSum] = stack.pop();
        visited[node] = true;

        let isLeaf = true;
        for (let neighbor of adj[node]) {
            if (!visited[neighbor]) {
                isLeaf = false;
                stack.push([neighbor, currentPathSum + values[neighbor]]);
            }
        }

        if (isLeaf && currentPathSum !== 0) {
            maxScore = Math.max(maxScore, currentPathSum);
        }
    }
};

let n = 6;
let edges = [[0, 1], [0, 2], [0, 3], [2, 4], [4, 5]];
let values = [5, 2, 5, 2, 1, 1];
let adj = Array.from({ length: n }, () => []);
for (let [a, b] of edges) {
    adj[a].push(b);
    adj[b].push(a);
}
let maxScore = 0;
dfsIterative(adj, values, n);
console.log(`Maximum Score: ${maxScore}`);
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

#define MAXN 20000

int values[MAXN];
int adj[MAXN][MAXN];
bool visited[MAXN];
int max_score;

struct Stack {
    int top;
    unsigned capacity;
    int* array;
};

struct Stack* createStack(unsigned capacity) {
    struct Stack* stack = (struct Stack*)malloc(sizeof(struct Stack));
    stack->capacity = capacity;
    stack->top = -1;
    stack->array = (int*)malloc(stack->capacity * sizeof(int));
    return stack;
}

bool isFull(struct Stack* stack) {
    return stack->top == stack->capacity - 1;
}

bool isEmpty(struct Stack* stack) {
    return stack->top == -1;
}

void push(struct Stack* stack, int item) {
    if (isFull(stack))
        return;
    stack->array[++stack->top] = item;
}

int pop(struct Stack* stack) {
    if (isEmpty(stack))
        return -1;
    return stack->array[stack->top--];
}

int peek(struct Stack* stack) {
    if (isEmpty(stack))
        return -1;
    return stack->array[stack->top];
}

void dfsIterative(int start, int n) {
    struct Stack* stack = createStack(n);
    int path_sum[n];
    push(stack, start);
    path_sum[start] = values[start];
    
    while (!isEmpty(stack)) {
        int node = pop(stack);
        visited[node] = true;
        
        bool isLeaf = true;

        for (int i = 0; i < n; i++) {
            if (adj[node][i] && !visited[i]) {
                isLeaf = false;
                path_sum[i] = path_sum[node] + values[i];
                push(stack, i);
            }
        }
        
        if (isLeaf && path_sum[node] != 0) {
            if (max_score < path_sum[node]) {
                max_score = path_sum[node];
            }
        }
    }
}

int main() {
    int n;
    // Example input
    int edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {2, 4}, {4, 5}};
    int values_in[] = {5, 2, 5, 2, 1, 1};
    n = 6;

    memcpy(values, values_in, sizeof(values_in));
    memset(visited, false, sizeof(visited));
    memset(adj, 0, sizeof(adj));

    for (int i = 0; i < n - 1; i++) {
        adj[edges[i][0]][edges[i][1]] = 1;
        adj[edges[i][1]][edges[i][0]] = 1;
    }

    max_score = 0;
    dfsIterative(0, n);

    printf("Maximum Score: %d\n", max_score);
    return 0;
}
using System;
using System.Collections.Generic;

class TreeMaxScore {
    static int maxScore;

    static void DFSIterative(List<int>[] adj, int[] values, int n) {
        bool[] visited = new bool[n];
        Stack<(int, int)> stack = new Stack<(int, int)>();
        stack.Push((0, values[0]));

        while (stack.Count > 0) {
            var (node, currentPathSum) = stack.Pop();
            visited[node] = true;

            bool isLeaf = true;

            foreach (var neighbor in adj[node]) {
                if (!visited[neighbor]) {
                    isLeaf = false;
                    stack.Push((neighbor, currentPathSum + values[neighbor]));
                }
            }

            if (isLeaf && currentPathSum != 0) {
                maxScore = Math.Max(maxScore, currentPathSum);
            }
        }
    }

    public static void Main() {
        int n = 6;
        int[][] edges = new int[][] { new int[]{0, 1}, new int[]{0, 2}, new int[]{0, 3}, new int[]{2, 4}, new int[]{4, 5} };
        int[] values = new int[] { 5, 2, 5, 2, 1, 1 };
        List<int>[] adj = new List<int>[n];
        for (int i = 0; i < n; i++) adj[i] = new List<int>();

        foreach (var edge in edges) {
            adj[edge[0]].Add(edge[1]);
            adj[edge[1]].Add(edge[0]);
        }

        maxScore = 0;
        DFSIterative(adj, values, n);

        Console.WriteLine("Maximum Score: " + maxScore);
    }
}

The Python implementation efficiently uses a stack data structure to perform DFS, recording the path score while ensuring every path ending at a leaf node contributes maximally if the path sum is not zero.

Complexity

Time Complexity: O(n) - Every node traversed once.
Space Complexity: O(n) - Stack facet utilized as data cache.

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Approach 3: Tree DP

The problem is actually asking us to select some nodes from all nodes of the tree so that the sum of these nodes' values is maximized, and there is one node on each path from the root node to the leaf node that is not selected.

We can use the method of tree DP to solve this problem.

We design a function dfs(i, fa), where i represents the current node with node i as the root of the subtree, and fa represents the parent node of i. The function returns an array of length 2, where [0] represents the sum of the values of all nodes in the subtree, and [1] represents the maximum value of the subtree satisfying that there is one node not selected on each path.

The value of [0] can be obtained directly by DFS accumulating the values of each node, while the value of [1] needs to consider two situations, namely whether node i is selected. If it is selected, then each subtree of node i must satisfy that there is one node not selected on each path; if it is not selected, then all nodes of each subtree of node i can be selected. We take the maximum of these two situations.

It should be noted that the value of [1] of the leaf node is 0, because the leaf node has no subtree, so there is no need to consider the situation where there is one node not selected on each path.

The answer is dfs(0, -1)[1].

The time complexity is O(n), and the space complexity is O(n). Here, n is the number of nodes.

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];
}
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
}
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Complexity comparison

ApproachComplexity
DFS and Tree Traversal

Time Complexity: O(n) - The tree is explored entirely once.
Space Complexity: O(n) - Recursive stack used and adjacency list utilized.

Iterative DFS using Stack

Time Complexity: O(n) - Every node traversed once.
Space Complexity: O(n) - Stack facet utilized as data cache.

Tree DP–

Complexity analysis

ApproachTimeSpaceWhen to Use
DFS with Tree Dynamic ProgrammingO(n)O(n)Best general solution. Clean recursive logic and optimal performance.
Iterative DFS using StackO(n)O(n)Useful when recursion depth may overflow or when iterative traversal is preferred.

Frequently asked questions

Is Maximum Score After Applying Operations on a Tree easy or hard?

The problem is typically classified as Medium difficulty. The challenge is recognizing that maximizing score is equivalent to minimizing the value that must remain in the tree. Once that insight is clear, a single DFS with tree dynamic programming solves it in linear time.

Maximum Score After Applying Operations on a Tree Python/Java solution

In Python or Java, construct the tree using an adjacency list and run a DFS that returns the required value for each subtree. Maintain the total sum of node values, compute subtree requirements with min(nodeValue, sum(childResults)), and subtract the root requirement from the total to get the maximum score.

How to solve Maximum Score After Applying Operations on a Tree in O(n)?

Build an adjacency list and run a post‑order DFS from the root. For leaf nodes, return their value since removing them would break the path constraint. For internal nodes, compute the sum of required values from children and return min(nodeValue, childSum). Track the total sum of all nodes and subtract the root’s required value to get the maximum score.

What is the best approach for Maximum Score After Applying Operations on a Tree?

The optimal approach uses DFS with tree dynamic programming. Traverse the tree bottom‑up and compute the minimum value that must remain in each subtree to keep every root‑to‑leaf path non‑zero. For each node, return min(nodeValue, sum of required values from children). The final score equals total node values minus the required value at the root. This runs in O(n) time.

Is Maximum Score After Applying Operations on a Tree asked at Google/Amazon/Meta?

Problems combining tree traversal and dynamic programming frequently appear in interviews at companies like Google, Amazon, and Meta. Variants that require computing optimal values along root‑to‑leaf paths or subtree decisions are common in senior‑level coding interviews.

What data structure is used in Maximum Score After Applying Operations on a Tree?

The solution uses an adjacency list to represent the tree and Depth‑First Search for traversal. The algorithm also applies dynamic programming on the tree structure, where each node stores the minimum value required to preserve valid root‑to‑leaf paths.

What is the time complexity of Maximum Score After Applying Operations on a Tree?

The optimal DFS solution runs in O(n) time where n is the number of nodes in the tree. Each node and edge is visited once during traversal. Space complexity is O(n) due to the adjacency list and recursion stack (or explicit stack in the iterative version).

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