[Leetcode][python] 690 - Employee importance - DFS

• Leetcode 690: Employee importance
• Level: easy
• Link

Employee Importance - LeetCode

My approach

1. Iterate through nested list and store employee id and importance in a dictionary
2. Perform dfs with starting id
   1. sum up the importance of an employee
   2. check if subordinate of the employee exists
      1. If exists, perform recursion on the subordinate employees
3. Return the total sum of employee

class Solution:    
        
    def getImportance(self, employees: List['Employee'], id: int) -> int:
        diction = {}
        
        # store employee id and importance
        for info in employees:
            diction[info.id] = info.importance
            
        res = 0 
        for employee in employees:
            if employee.id == id:
                res += employee.importance
                if employee.subordinates:
                    for subs in employee.subordinates:
                        res += self.getImportance(employees, subs)
        return (res)