leetcode 690. Employee Importance——本质上就是tree的DFS和BFS

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:

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

经典BFS：tree的层序遍历思想

"""
# Employee info
class Employee(object):
    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 Solution(object):
    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """
        """
        Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
        Output: 11
        use dict to find its sub employee
        """
        emp_dict = {e.id:e for i,e in enumerate(employees)}
        root = emp_dict[id]
        # tree BFS
        q = [root]
        ans = 0
        while q:
            q2 = []
            for node in q:
                ans += node.importance
                for i in node.subordinates:
                    q2.append(emp_dict[i])
            q = q2
        return ans        

DFS:

"""
# Employee info
class Employee(object):
    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 Solution(object):
    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """
        """
        Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
        Output: 11
        use dict to find its sub employee
        """
        emp_dict = {e.id:e for i,e in enumerate(employees)}
        root = emp_dict[id]

        # tree DFS
        def dfs(root, emp_dict):
            score = root.importance
            for i in root.subordinates:
                score += dfs(emp_dict[i], emp_dict)
            return score

        return dfs(root, emp_dict)