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.
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)