There is a binary tree rooted at 0
consisting of n
nodes. The nodes are labeled from 0
to n - 1
. You are given a 0-indexed integer array parents
representing the tree, where parents[i]
is the parent of node i
. Since node 0
is the root, parents[0] == -1
.
Each node has a score. To find the score of a node, consider if the node and the edges connected to it were removed. The tree would become one or more non-empty subtrees. The size of a subtree is the number of the nodes in it. The score of the node is the product of the sizes of all those subtrees.
Return the number of nodes that have the highest score.
Example 1:
Input: parents = [-1,2,0,2,0] Output: 3 Explanation: - The score of node 0 is: 3 * 1 = 3 - The score of node 1 is: 4 = 4 - The score of node 2 is: 1 * 1 * 2 = 2 - The score of node 3 is: 4 = 4 - The score of node 4 is: 4 = 4 The highest score is 4, and three nodes (node 1, node 3, and node 4) have the highest score.
Example 2:
Input: parents = [-1,2,0] Output: 2 Explanation: - The score of node 0 is: 2 = 2 - The score of node 1 is: 2 = 2 - The score of node 2 is: 1 * 1 = 1 The highest score is 2, and two nodes (node 0 and node 1) have the highest score.
Constraints:
n == parents.length
2 <= n <= 105
parents[0] == -1
0 <= parents[i] <= n - 1
fori != 0
parents
represents a valid binary tree.
class Solution:
def countHighestScoreNodes(self, parents: List[int]) -> int:
n, max_score, ans = len(parents), 0, 0
g = [[] for _ in range(n)]
for i in range(1, n):
g[parents[i]].append(i)
def dfs(cur: int) -> int:
nonlocal max_score, ans
size, score = 1, 1
for c in g[cur]:
s = dfs(c)
size += s
score *= s
if cur > 0:
score *= n - size
if score > max_score:
max_score = score
ans = 1
elif score == max_score:
ans += 1
return size
dfs(0)
return ans
class Solution {
private int n;
private long maxScore;
private int ans;
private List<List<Integer>> graph;
public int countHighestScoreNodes(int[] parents) {
n = parents.length;
maxScore = 0;
ans = 0;
graph = new ArrayList<>();
for (int i = 0; i < n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 1; i < n; i++) {
graph.get(parents[i]).add(i);
}
dfs(0);
return ans;
}
private int dfs(int cur) {
int size = 1;
long score = 1;
for (int child : graph.get(cur)) {
int s = dfs(child);
size += s;
score *= s;
}
if (cur > 0) {
score *= n - size;
}
if (score > maxScore) {
maxScore = score;
ans = 1;
} else if (score == maxScore) {
ans++;
}
return size;
}
}
function countHighestScoreNodes(parents: number[]): number {
const n = parents.length;
let edge = Array.from({ length: n }, (v, i) => []);
for (let i = 0; i < n; i++) {
const parent = parents[i];
if (parent != -1) {
edge[parent].push(i);
}
}
let ans = 0;
let max = 0;
function dfs(idx: number): number {
let size = 1,
score = 1;
for (let i = 0; i < edge[idx].length; i++) {
const child = edge[idx][i];
let childSize = dfs(child);
size += childSize;
score *= childSize;
}
if (idx > 0) {
score *= n - size;
}
if (score > max) {
max = score;
ans = 1;
} else if (score == max) {
ans++;
}
return size;
}
dfs(0);
return ans;
}
class Solution {
public:
int ans;
long long maxScore;
int n;
int countHighestScoreNodes(vector<int>& parents) {
ans = 0;
maxScore = 0;
n = parents.size();
unordered_map<int, vector<int>> g;
for (int i = 1; i < n; ++i) g[parents[i]].push_back(i);
dfs(0, g);
return ans;
}
int dfs(int u, unordered_map<int, vector<int>>& g) {
int size = 1;
long long score = 1;
for (int v : g[u]) {
int t = dfs(v, g);
size += t;
score *= t;
}
if (u > 0) score *= (n - size);
if (score > maxScore) {
maxScore = score;
ans = 1;
} else if (score == maxScore)
++ans;
return size;
}
};
func countHighestScoreNodes(parents []int) int {
n := len(parents)
g := make([][]int, n)
for i := 1; i < n; i++ {
p := parents[i]
g[p] = append(g[p], i)
}
maxScore, ans := 0, 0
var dfs func(int) int
dfs = func(u int) int {
size, score := 1, 1
for _, v := range g[u] {
t := dfs(v)
size += t
score *= t
}
if u > 0 {
score *= n - size
}
if score > maxScore {
maxScore, ans = score, 1
} else if score == maxScore {
ans++
}
return size
}
dfs(0)
return ans
}