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

**The Idea:** DFS and accumulate the graph weights.

**Complexity:** O(N) time and space

```python
# Employee info
class Employee:
    def __init__(self, id, importance, subordinates):
        # It's the unique id of each node.
        # unique id of this employee
        self.id = id
        # the importance value of this employee
        self.importance = importance
        # the id of direct subordinates
        self.subordinates = subordinates


def getImportance(self, employees, id):
    """
    :type employees: Employee
    :type id: int
    :rtype: int
    """

    graph = {employee.id:{'weight':employee.importance, 'subordinates':employee.subordinates}
             for employee in employees}
    visited = set()

    def dfs(root):
        visited.add(root)
        subordinates = graph[root]['subordinates']
        if not subordinates:
            return graph[root]['weight']
        else:
            sum = graph[root]['weight']
            for employee in subordinates:
                if employee not in visited:
                    sum += dfs(employee)
            return sum

    return dfs(id)
```
