# 1443. Minimum Time to Collect All Apples in a Tree

## Description

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.

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 <= ai < bi <= n - 1
• hasApple.length == n

## Solutions

DFS.

• 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];
}
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;
}
}

• class Solution {
public:
int minTime(int n, vector<vector<int>>& edges, vector<bool>& hasApple) {
vector<bool> vis(n);
vector<vector<int>> g(n);
for (auto& e : edges) {
int u = e[0], v = e[1];
g[u].push_back(v);
g[v].push_back(u);
}
return dfs(0, 0, g, hasApple, vis);
}

int dfs(int u, int cost, vector<vector<int>>& g, vector<bool>& hasApple, vector<bool>& vis) {
if (vis[u]) return 0;
vis[u] = true;
int nxt = 0;
for (int& v : g[u]) nxt += dfs(v, 2, g, hasApple, vis);
if (!hasApple[u] && !nxt) return 0;
return cost + nxt;
}
};

• 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)


• 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)
}