贴一个我的，思路也是用邻接表保存节点的string和他自身是哪一个节点，然后串在每个父节点后面最后dfs。就是前缀的那些符号条件要仔细考虑。

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

void DFS(vector<vector<pair<string,int>>> &dir, int root, string mask) {
    string next_mask;
    for (int i = 0; i < dir[root].size(); ++i) {
        pair<string, int> node = dir[root][i];
        cout << mask;
        if (i != dir[root].size() - 1) {
            cout << "|-- ";
        }
        else {
            cout << "`-- ";
        }
        cout << node.first << endl;

        if (i == dir[root].size() - 1)
        {
            next_mask = "    " + mask;
        }
        else {
            next_mask = mask + "|   ";
        }
        DFS(dir, node.second, next_mask);
    }
}

bool cmp(pair<string, int> a, pair<string, int> b) {
    return a.first < b.first;
}

int main() {
    int n;
    cin >> n;
    vector<vector<pair<string,int>>> dir(n + 1);
    for (int i = 0; i < n; ++i) {
        string node_s;
        int fa_idx;
        pair<string, int> node;
        cin >> node_s >> fa_idx;
        node.first = node_s;
        node.second = i+1;
        dir[fa_idx + 1].push_back(node);
    }

    for (int i = 0; i < dir.size(); ++i) {
        sort(dir[i].begin(), dir[i].end(),cmp);
    }

    cout << dir[0][0].first << endl;
    DFS(dir, 1, "");
    return 0;
}