582. Kill Process

问题描述：

Given n processes, each process has a unique PID (process id) and its PPID (parent process id).

Each process only has one parent process, but may have one or more children processes. This is just like a tree structure. Only one process has PPID that is 0, which means this process has no parent process. All the PIDs will be distinct positive integers.

We use two list of integers to represent a list of processes, where the first list contains PID for each process and the second list contains the corresponding PPID.

Now given the two lists, and a PID representing a process you want to kill, return a list of PIDs of processes that will be killed in the end. You should assume that when a process is killed, all its children processes will be killed. No order is required for the final answer.

Example 1:

Input: 
pid =  [1, 3, 10, 5]
ppid = [3, 0, 5, 3]
kill = 5
Output: [5,10]
Explanation: 
           3
         /   
        1     5
             /
            10
Kill 5 will also kill 10.

Note:

1. The given kill id is guaranteed to be one of the given PIDs.
2. n >= 1.

解题思路：

容易想到：首先用一个unordered_map<int, vector<int>> 来存储父节点，与其可能的直系子节点。

然后在map中寻找kill的左右子孙节点。

这里用queue来辅助我们进行遍历。

代码：

class Solution {
public:
    vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {
        vector<int> ret;
        unordered_map<int, vector<int>> m;
        for(int i = 0; i < pid.size(); i++){
            m[ppid[i]].push_back(pid[i]);
        }
        
        if(m.count(kill) == 0){
            ret.push_back(kill);
            return ret;
        } 
        
        queue<int> q;
        q.push(kill);
        while(!q.empty()){
            int p = q.front();
            ret.push_back(p);
            q.pop();
            if(m.count(p)){
                for(int i : m[p]){
                    q.push(i);
                }
            }
        }
        
        return ret;
    }
};