leetcode 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 importan

Note:
One employee has at most one direct leader and may have several subordinates.
The maximum number of employees won't exceed 2000.

按照要求，递归遍历，对节点求和。

/*
// 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;
};
*/
class Solution {
public:
    int sum = 0;
    unordered_map<int, int> mp;
    void dfs(vector<int>& v, vector<Employee*>& e) {
        for (int i = 0; i < v.size(); ++i) {
            if (e[mp[v[i]]]->subordinates.size()) {
                dfs(e[mp[v[i]]]->subordinates, e);
            }
            sum +=  e[mp[v[i]]]->importance;
        }
    }
    int getImportance(vector<Employee*> employees, int id) {
        if (employees.size() == 0) return 0;
        for (int i = 0; i < employees.size(); ++i) {
            mp[employees[i]->id] = i;
        }
        vector<int> v;
        sum += employees[mp[id]]->importance;
        v = employees[mp[id]]->subordinates;
        if (v.size())dfs(v, employees);
        return sum;
    }
};