Employee Importance LeetCode solution in Python

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# Runtime: 254 ms (Top 40.58%) | Memory: 15.6 MB (Top 60.31%)
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"""
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# Definition for Employee.
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class Employee:
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    def __init__(self, id: int, importance: int, subordinates: List[int]):
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        self.id = id
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        self.importance = importance
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        self.subordinates = subordinates
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"""
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class Solution:
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    def dfs(self, graph, employees, empId, totalImportance, visited):
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        totalImportance += graph[empId][0]
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        visited.add(empId)
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        for neighbour in graph[empId][1]:
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            if neighbour not in visited:
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                totalImportance = self.dfs(
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                    graph, employees, neighbour, totalImportance, visited
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                )
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        return totalImportance
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    def getImportance(self, employees: List["Employee"], id: int) -> int:
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        graph = {}
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        for employeeInfo in employees:
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            empId, importance, suboordinates = (
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                employeeInfo.id,
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                employeeInfo.importance,
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                employeeInfo.subordinates,
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            )
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            graph[empId] = [importance, suboordinates]
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        n = len(employees)
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        if graph[id][1] == []:
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            return graph[id][0]
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        visited = set()
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        totalImportance = 0
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        totalImportance = self.dfs(graph, employees, id, totalImportance, visited)
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        return totalImportance