leetcode 690. Employee Importance 深度优先遍历DFS

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.

本题题意很简单，就是做一个递归搜索即可

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>#include <regex>using namespace std;/*// Employee infoclass Employee {    public:    // It's the unique ID of each node.    // unique id of this employee    int id;    // the importance value of this employee    int importance;    // the id of direct subordinates    vector<int> subordinates;};*/class Solution {public:    int getImportance(vector<Employee*> employees, int id)     {        if (employees.size() <= 0)            return 0;        else        {            Employee* tar = NULL;            for (auto i : employees)            {                if (i->id == id)                {                    tar = i;                    break;                }            }            int sum = tar->importance;            for (int next : tar->subordinates)            {                sum += getImportance(employees, next);            }            return sum;        }    }};