690. Employee Importance(DPS，员工价值树)

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.

// 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;
};*/

class Solution {

    public int getImportance(List<Employee> employees, int id) {
        Map<Integer, Employee> emap = new HashMap<>();
        for (Employee person : employees) emap.put(person.id, person);
        return ret(emap,id);
    }

    public static int ret(Map<Integer,Employee> emap,int id){
        int sum=0;
        Employee target = emap.get(id);
        sum += target.importance;
        for (Integer item : target.subordinates) {
            sum+=ret(emap,item);
        }
        return sum;
    }
}