Employee Importance

Problem

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:

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

Solution

/*
// Employee info
class Employee {
    // It's the unique id of each node;
    // unique id of this employee
    public int id;
    // the importance value of this employee
    public int importance;
    // the id of direct subordinates
    public List<Integer> subordinates;
};
*/

Recursive Solution

public class Solution {
    public int getImportance(List<Employee> employees, int id) {
        if (employees == null || employees.size() == 0) return 0;
        Employee manager = null;
        for (Employee e : employees) {
            if (e.id == id) {
                manager = e;
                break;
            }
        }
        int result = manager.importance;
        for (int i : manager.subordinates) result += getImportance(employees, i);
        return result;
    }
}

Analysis

At first I thought about using a List<Integer> to collect all subordinates' ids
And update it by calling list.addAll(e.subordinates)
This won't work because if a subordinate is behind the current employee
It's better we use recursive and update result by calling method again with id in manager.subordinates