LeetCode #1443 — MEDIUM

Minimum Time to Collect All Apples in a Tree

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

Given an undirected tree consisting of n vertices numbered from 0 to n-1, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. Return the minimum time in seconds you have to spend to collect all apples in the tree, starting at vertex 0 and coming back to this vertex.

The edges of the undirected tree are given in the array edges, where edges[i] = [a_i, b_i] means that exists an edge connecting the vertices a_i and b_i. Additionally, there is a boolean array hasApple, where hasApple[i] = true means that vertex i has an apple; otherwise, it does not have any apple.

Example 1:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,true,true,false]
Output: 8 
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.

Example 2:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,false,true,false]
Output: 6
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.

Example 3:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,false,false,false,false,false]
Output: 0

Constraints:

1 <= n <= 10⁵
edges.length == n - 1
edges[i].length == 2
0 <= a_i < b_i <= n - 1
hasApple.length == n

Step 01

Brute Force Baseline

Problem summary: Given an undirected tree consisting of n vertices numbered from 0 to n-1, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. Return the minimum time in seconds you have to spend to collect all apples in the tree, starting at vertex 0 and coming back to this vertex. The edges of the undirected tree are given in the array edges, where edges[i] = [ai, bi] means that exists an edge connecting the vertices ai and bi. Additionally, there is a boolean array hasApple, where hasApple[i] = true means that vertex i has an apple; otherwise, it does not have any apple.

Baseline thinking

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

Pattern signal: Hash Map · Tree

Example 1

7
[[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]]
[false,false,true,false,true,true,false]

Example 2

7
[[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]]
[false,false,true,false,false,true,false]

Example 3

7
[[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]]
[false,false,false,false,false,false,false]

Step 02

Core Insight

What unlocks the optimal approach

Note that if a node u contains an apple then all edges in the path from the root to the node u have to be used forward and backward (2 times).
Therefore use a depth-first search (DFS) to check if an edge will be used or not.

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 #1443: Minimum Time to Collect All Apples in a Tree
class Solution {
    public int minTime(int n, int[][] edges, List<Boolean> hasApple) {
        boolean[] vis = new boolean[n];
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] e : edges) {
            int u = e[0], v = e[1];
            g[u].add(v);
            g[v].add(u);
        }
        return dfs(0, 0, g, hasApple, vis);
    }

    private int dfs(int u, int cost, List<Integer>[] g, List<Boolean> hasApple, boolean[] vis) {
        if (vis[u]) {
            return 0;
        }
        vis[u] = true;
        int nxtCost = 0;
        for (int v : g[u]) {
            nxtCost += dfs(v, 2, g, hasApple, vis);
        }
        if (!hasApple.get(u) && nxtCost == 0) {
            return 0;
        }
        return cost + nxtCost;
    }
}

// Accepted solution for LeetCode #1443: Minimum Time to Collect All Apples in a Tree
func minTime(n int, edges [][]int, hasApple []bool) int {
	vis := make([]bool, n)
	g := make([][]int, n)
	for _, e := range edges {
		u, v := e[0], e[1]
		g[u] = append(g[u], v)
		g[v] = append(g[v], u)
	}
	var dfs func(int, int) int
	dfs = func(u, cost int) int {
		if vis[u] {
			return 0
		}
		vis[u] = true
		nxt := 0
		for _, v := range g[u] {
			nxt += dfs(v, 2)
		}
		if !hasApple[u] && nxt == 0 {
			return 0
		}
		return cost + nxt
	}
	return dfs(0, 0)
}

# Accepted solution for LeetCode #1443: Minimum Time to Collect All Apples in a Tree
class Solution:
    def minTime(self, n: int, edges: List[List[int]], hasApple: List[bool]) -> int:
        def dfs(u, cost):
            if vis[u]:
                return 0
            vis[u] = True
            nxt_cost = 0
            for v in g[u]:
                nxt_cost += dfs(v, 2)
            if not hasApple[u] and nxt_cost == 0:
                return 0
            return cost + nxt_cost

        g = defaultdict(list)
        for u, v in edges:
            g[u].append(v)
            g[v].append(u)
        vis = [False] * n
        return dfs(0, 0)

// Accepted solution for LeetCode #1443: Minimum Time to Collect All Apples in a Tree
use std::collections::HashMap;

impl Solution {
    pub fn min_time(n: i32, edges: Vec<Vec<i32>>, has_apple: Vec<bool>) -> i32 {
        let mut adj = HashMap::new();

        for edge in edges {
            let (parent, child) = (edge[0], edge[1]);
            adj.entry(parent).or_insert(vec![]).push(child);
            adj.entry(child).or_insert(vec![]).push(parent);
        }

        fn dfs(
            current: i32,
            parent: i32,
            adj: &HashMap<i32, Vec<i32>>,
            has_apple: &Vec<bool>,
        ) -> i32 {
            let mut time = 0;

            if let Some(children) = adj.get(&current) {
                for child in children {
                    if *child == parent {
                        continue;
                    }
                    let child_time = dfs(*child, current, &adj, &has_apple);

                    if child_time.is_positive() || has_apple[*child as usize] {
                        time += 2 + child_time;
                    }
                }
            }

            time
        }

        dfs(0, -1, &adj, &has_apple)
    }
}

// Accepted solution for LeetCode #1443: Minimum Time to Collect All Apples in a Tree
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1443: Minimum Time to Collect All Apples in a Tree
// class Solution {
//     public int minTime(int n, int[][] edges, List<Boolean> hasApple) {
//         boolean[] vis = new boolean[n];
//         List<Integer>[] g = new List[n];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (int[] e : edges) {
//             int u = e[0], v = e[1];
//             g[u].add(v);
//             g[v].add(u);
//         }
//         return dfs(0, 0, g, hasApple, vis);
//     }
// 
//     private int dfs(int u, int cost, List<Integer>[] g, List<Boolean> hasApple, boolean[] vis) {
//         if (vis[u]) {
//             return 0;
//         }
//         vis[u] = true;
//         int nxtCost = 0;
//         for (int v : g[u]) {
//             nxtCost += dfs(v, 2, g, hasApple, vis);
//         }
//         if (!hasApple.get(u) && nxtCost == 0) {
//             return 0;
//         }
//         return cost + nxtCost;
//     }
// }

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(h)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.