統計每個子樹的個數就行

當前節點的value = 它的所有子樹的節點個數 + 根節點樹節點的個數 - 當前節點子樹節點的個數

class Solution {
public:
    vector<int> v[100010];
    int num[100010];
    long long ret[100010];
    int cnt;

    int dfs(int u)
    {
        int len = v[u].size();
        int cnt1 = 0, cnt2 = 0;
        for(int i = 0; i < len; i++)
        {
            
 
if(i == 0)
                cnt1 = dfs(v[u][i]);
            else
                cnt2 = dfs(v[u][i]);
        }
        num[u] = cnt1 + cnt2 + 1;
        return num[u];
    }
    
    
    
    
    
    int countHighestScoreNodes(vector<int>& parents) {
        int n = parents.size();
        
 
for(int i = 0; i < n; i++)
            if(i != 0) v[parents[i]].push_back(i);
        int k = dfs(0);
        long long max_v = -1, sum;
        
        for(int i = 0; i < n; i++)
        {
            sum = 1;
            if(i == 0)
            {
                int len2 = v[i].size();
                
 
for(int j = 0; j < len2; j++)
                    sum *= (long long)num[v[i][j]];
            }
            else
            {
                int len2 = v[i].size();
                sum *= (long long)(num[0] - num[i]);
                for(int j = 0; j < len2; j++)
                    sum *= (long long)num[v[i][j]];
            }
            ret[i] = sum;
            max_v = max(max_v, sum);
            
        }
        int ans = 0;
        for(int i = 0; i < n; i++)
            if(ret[i] == max_v) ans++;
        
        
        return ans;
        
        
    }
};