[leetcode]Breadth-first Search-690. Employee Importance

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

class Solution
{
    public int getImportance(List employees, int id)
    {
        int index = 0;
        for(index = 0; index < employees.size(); index++)
        {
            if(employees.get(index).id == id)
                break;
        }
        int result = employees.get(index).importance;
        for(int i = 0; i < employees.get(index).subordinates.size(); i++)
        {
            result += getImportance(employees, employees.get(index).subordinates.get(i));
        }
        return result;
    }
}