leetcode 582.Kill Process

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.

很自然的想法，把他建成一棵树，然后依次加入子孙，于是有了DFS和BFS两种想法。
注意：set, map, vector等常用STL的用法
先是dfs

void killID(map<int, set<int>>&v, vector<int> &killed, int killid)    {           //递归的想法很常见，也很重要    killed.push_back(killid);    for (int child : v[killid])        killID(v, killed, child);    }    vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {    map<int, set<int>> children;    vector<int> killed;    for (int i = 0; i < ppid.size(); i++)        children[ppid[i]].insert(pid[i]);    killID(children, killed, kill);    return killed;    }

BFS:

vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {    map<int, set<int>> children;    vector<int> killed;    queue<int> child;    for (int i = 0; i<ppid.size(); i++)        children[ppid[i]].insert(pid[i]);    child.push(kill);    while (!child.empty())    {        int a = child.front();                             child.pop();//出队        killed.push_back(a);//记录        for (int i : children[a])//遍历            child.push(i);    }    return killed;    }