RT，做法是先从 $1$ 开始，找另外字符串（未访问过）的最大公共前缀的最大值，然后递归并输出

int f(string s1, string s2) {//最大公共前缀
    int ret = 0;
    for (int i = 0; i < min(s1.size(), s2.size()); ++i) {
        if (s1[i] == s2[i]) ++ret;
        else return ret;
    }
}

string s[9961];

void NTT(int p) {
    int mx = -0x3f3f3f3f, w = 0;
    cout << s[p] << '\n';
    vis[p] = 1;
    for (int i = 1; i <= n; ++i) {
        int toptree = f(s[p], s[i]);
        if (toptree > mx && !vis[i]) {
            mx = toptree;
            w = i;
        } 
    }
    if (w) NTT(w);
}