690. Employee Importance

You are given a data structure of employee information, which includes the employee's unique id, his importance value and his direct subordinates' id.

For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct.

Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all his subordinates.

Example 1:

Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
Output: 11
Explanation:
Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3. So the total importance value of employee 1 is 5 + 3 + 3 = 11.

Note:

1. One employee has at most one direct leader and may have several subordinates.

2. The maximum number of employees won't exceed 2000.

// Employee info
class Employee {
public:
    // It's the unique ID of each node.
    // unique id of this employee
    int id;
    // the importance value of this employee
    int importance;
    // the id of direct subordinates
    vector<int> subordinates;
};

// Recursive DFS
void helper(unordered_map<int, Employee*>& mp, int id, int& res) {
    if (!mp.count(id)) return;
    Employee*& emp = mp[id];
    for (int sub_id : emp->subordinates) helper(mp, sub_id, res);
    res += emp->importance;
}
int getImportance(vector<Employee*> employees, int id) {
    unordered_map<int, Employee*> mp; // id -> emp info
    for (Employee*& emp : employees) mp[emp->id] = emp;
    int res = 0;
    helper(mp, id, res);
    return res;
}

// BFS
int getImportance(vector<Employee*> employees, int id) {
    unordered_map<int, Employee*> mp; // id -> emp info
    for (Employee*& emp : employees) mp[emp->id] = emp;
    queue<int> q({id});
    int res = 0;
    while (!q.empty()) {
        Employee* t = mp[q.front()]; q.pop();
        res += t->importance;
        for (int sub_id : t->subordinates)
            q.push(sub_id);
    }
    return res;
}