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Formatted question description: https://leetcode.ca/all/2538.html
2538. Difference Between Maximum and Minimum Price Sum
Description
There exists an undirected and initially unrooted tree with n
nodes indexed from 0
to n  1
. You are given the integer n
and 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.
Each node has an associated price. You are given an integer array price
, where price[i]
is the price of the i^{th}
node.
The price sum of a given path is the sum of the prices of all nodes lying on that path.
The tree can be rooted at any node root
of your choice. The incurred cost after choosing root
is the difference between the maximum and minimum price sum amongst all paths starting at root
.
Return the maximum possible cost amongst all possible root choices.
Example 1:
Input: n = 6, edges = [[0,1],[1,2],[1,3],[3,4],[3,5]], price = [9,8,7,6,10,5] Output: 24 Explanation: The diagram above denotes the tree after rooting it at node 2. The first part (colored in red) shows the path with the maximum price sum. The second part (colored in blue) shows the path with the minimum price sum.  The first path contains nodes [2,1,3,4]: the prices are [7,8,6,10], and the sum of the prices is 31.  The second path contains the node [2] with the price [7]. The difference between the maximum and minimum price sum is 24. It can be proved that 24 is the maximum cost.
Example 2:
Input: n = 3, edges = [[0,1],[1,2]], price = [1,1,1] Output: 2 Explanation: The diagram above denotes the tree after rooting it at node 0. The first part (colored in red) shows the path with the maximum price sum. The second part (colored in blue) shows the path with the minimum price sum.  The first path contains nodes [0,1,2]: the prices are [1,1,1], and the sum of the prices is 3.  The second path contains node [0] with a price [1]. The difference between the maximum and minimum price sum is 2. It can be proved that 2 is the maximum cost.
Constraints:
1 <= n <= 10^{5}
edges.length == n  1
0 <= a_{i}, b_{i} <= n  1
edges
represents a valid tree.price.length == n
1 <= price[i] <= 10^{5}
Solutions

class Solution { private List<Integer>[] g; private long ans; private int[] price; public long maxOutput(int n, int[][] edges, int[] price) { g = new List[n]; 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); } this.price = price; dfs(0, 1); return ans; } private long[] dfs(int i, int fa) { long a = price[i], b = 0; for (int j : g[i]) { if (j != fa) { var e = dfs(j, i); long c = e[0], d = e[1]; ans = Math.max(ans, Math.max(a + d, b + c)); a = Math.max(a, price[i] + c); b = Math.max(b, price[i] + d); } } return new long[] {a, b}; } }

class Solution { public: long long maxOutput(int n, vector<vector<int>>& edges, vector<int>& price) { vector<vector<int>> g(n); for (auto& e : edges) { int a = e[0], b = e[1]; g[a].push_back(b); g[b].push_back(a); } using ll = long long; using pll = pair<ll, ll>; ll ans = 0; function<pll(int, int)> dfs = [&](int i, int fa) { ll a = price[i], b = 0; for (int j : g[i]) { if (j != fa) { auto [c, d] = dfs(j, i); ans = max({ans, a + d, b + c}); a = max(a, price[i] + c); b = max(b, price[i] + d); } } return pll{a, b}; }; dfs(0, 1); return ans; } };

class Solution: def maxOutput(self, n: int, edges: List[List[int]], price: List[int]) > int: def dfs(i, fa): a, b = price[i], 0 for j in g[i]: if j != fa: c, d = dfs(j, i) nonlocal ans ans = max(ans, a + d, b + c) a = max(a, price[i] + c) b = max(b, price[i] + d) return a, b g = defaultdict(list) for a, b in edges: g[a].append(b) g[b].append(a) ans = 0 dfs(0, 1) return ans

func maxOutput(n int, edges [][]int, price []int) int64 { g := make([][]int, n) for _, e := range edges { a, b := e[0], e[1] g[a] = append(g[a], b) g[b] = append(g[b], a) } type pair struct{ a, b int } ans := 0 var dfs func(i, fa int) pair dfs = func(i, fa int) pair { a, b := price[i], 0 for _, j := range g[i] { if j != fa { e := dfs(j, i) c, d := e.a, e.b ans = max(ans, max(a+d, b+c)) a = max(a, price[i]+c) b = max(b, price[i]+d) } } return pair{a, b} } dfs(0, 1) return int64(ans) } func max(a, b int) int { if a > b { return a } return b }