【Leetcode】690. Employee Importance

# 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

class Solution1:
    def getImportance(self, employees, id):
        """
        use dict , key is id , value is employee object, use id to locate an employee
        and use DFS to get all subordinates
        """
        dictionary = {}
        for employee in employees:
            dictionary[employee.id] = employee
        result = 0
        from collections import  deque
        queue = deque()
        queue.append(dictionary[id])
        while(len(queue)):
            top = queue.popleft()
            result += top.importance
            for sub in top.subordinates:
                queue.append(dictionary[sub])
        return result

class Solution2:
    """
    make code more concise
    """
    def getImportance(self, employees, id):
        dictionary = {employee.id:employee for employee in employees}
        def dfs(id):
            sub_importance = sum([dfs(sub_id) for sub_id in dictionary[id].subordinates])
            return sub_importance + dictionary[id].importance
        return dfs(id)

class Solution3:
    """
    make code more concise
    """
    def getImportance(self, employees, id):
        dictionary = {employee.id: employee for employee in employees}
        dfs = lambda id: sum([dfs(sub_id) for sub_id in dictionary[id].subordinates]) + dictionary[id].importance
        return dfs(id)