Problem:
Rooted Trees

A graph G = (V, E) is a data structure where V is a finite set of vertices and E is a binary relation on V represented by a set of edges. Fig. 1 illustrates an example of a graph (or graphs).

Fig. 1

A free tree is a connnected, acyclic, undirected graph. A rooted tree is a free tree in which one of the vertices is distinguished from the others. A vertex of a rooted tree is called "node."

Your task is to write a program which reports the following information for each node u of a given rooted tree T:

node ID of u

parent of u

depth of u

node type (root, internal node or leaf)

a list of chidlren of u

If the last edge on the path from the root r of a tree T to a node x is (p, x), then p is the parent of x, and x is a child of p. The root is the only node in T with no parent.

A node with no children is an external node or leaf. A nonleaf node is an internal node

The number of children of a node x in a rooted tree T is called the degree of x.

The length of the path from the root r to a node x is the depth of x in T.

Here, the given tree consists of n nodes and evey node has a unique ID from 0 to n-1.

Fig. 2 shows an example of rooted trees where ID of each node is indicated by a number in a circle (node). The example corresponds to the first sample input.

Fig. 2

Input

The first line of the input includes an integer n, the number of nodes of the tree.

In the next n lines, the information of each node u is given in the following format:

id k c1 c2 ... ck

where id is the node ID of u, k is the degree of u, c1 ... ck are node IDs of 1st, ... kth child of u. If the node does not have a child, the k is 0.

Output

Print the information of each node in the following format ordered by IDs:

node id: parent = p, depth = d, type, [c1...ck]

p is ID of its parent. If the node does not have a parent, print -1.

d is depth of the node.

type is a type of nodes represented by a string (root, internal node or leaf). If the root can be considered as a leaf or an internal node, print root.

c1...ck is the list of children as a ordered tree.

Please follow the format presented in a sample output below.

Constraints

1 ≤ n ≤ 100000

Sample Input 1

13
0 3 1 4 10
1 2 2 3
2 0
3 0
4 3 5 6 7
5 0
6 0
7 2 8 9
8 0
9 0
10 2 11 12
11 0
12 0

Sample Output 1

node 0: parent = -1, depth = 0, root, [1, 4, 10]
node 1: parent = 0, depth = 1, internal node, [2, 3]
node 2: parent = 1, depth = 2, leaf, []
node 3: parent = 1, depth = 2, leaf, []
node 4: parent = 0, depth = 1, internal node, [5, 6, 7]
node 5: parent = 4, depth = 2, leaf, []
node 6: parent = 4, depth = 2, leaf, []
node 7: parent = 4, depth = 2, internal node, [8, 9]
node 8: parent = 7, depth = 3, leaf, []
node 9: parent = 7, depth = 3, leaf, []
node 10: parent = 0, depth = 1, internal node, [11, 12]
node 11: parent = 10, depth = 2, leaf, []
node 12: parent = 10, depth = 2, leaf, []

Sample Input 2

4
1 3 3 2 0
0 0
3 0
2 0

Sample Output 2

node 0: parent = 1, depth = 1, leaf, []
node 1: parent = -1, depth = 0, root, [3, 2, 0]
node 2: parent = 1, depth = 1, leaf, []
node 3: parent = 1, depth = 1, leaf, []

Note

You can use a left-child, right-sibling representation to implement a tree which has the following data:

the parent of u

the leftmost child of u

the immediate right sibling of u

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C# (judge verdict: Accepted)

//根付き木
//できなかったとこ：入力数が行ごとに違うのでどう配列に入れるか
//解決策：構造体で作って項目が変化するところをLinq
//参考：https://onlinejudge.u-aizu.ac.jp/solutions/problem/ALDS1_7_A/review/1911082/emine/C%23
//
using System;
using System.Text;
using System.Collections.Generic;
namespace tst
{
    class Program
    {
        class Node
        {
            internal int parent;
            internal int depth;
            internal string type;
            internal List<int> child;//名前空間はSystem.Collections.Generic
            public Node()
            {
                parent = -1;
                depth = 0;
                type = "";
                child = new List<int>();
            }
        }
        
        public static void Main()
        {
            int n = int.Parse(Console.ReadLine());
            //まずクラスで[int parent, int depth, string type, List<int>chid]をつくりそこに保存
            Node[] nodes = new Node[n];
            //初期化//構造体だと初期化できない
            for(int i=0; i<n; i++)
            {
                nodes[i] = new Node();
            }
            //入力
            for(int i=0; i<n; i++)
            {
                string[] In = Console.ReadLine().Split(' ');
                //0 3 1 4 10だったら0のchildに1 4 10、1 4 10のparentに0を入れる、
                int k = int.Parse(In[1]);
                if (k>=1){
                    for(int j=2; j<= k+ 1; j++)
                    {
                        nodes[int.Parse(In[0])].child.Add(int.Parse(In[j]));//j,j+1,,,k+1をListの後ろに追加していく//勝手にnode順に並ぶ
                        nodes[int.Parse(In[j])].parent = int.Parse(In[0]);
                    }
                }
            }
            Stack<Node> stk = new Stack<Node>();//名前空間はSystem.Collections//後入れ先しのstack
            //typeを埋めていく
            for (int i=0; i<n; i++)
            {
                string mes = ""; 
                if (nodes[i].parent == -1)
                {
                    mes = "root";
                    stk.Push(nodes[i]);//rootをスタックの先頭に入れる
                }
                else if (nodes[i].child.Count >=1)
                {
                    mes = "internal node";
                }
                else if (nodes[i].child.Count ==0)
                {
                    mes = "leaf";
                }
                nodes[i].type = mes;
            }
            while (stk.Count > 0)//スタックに要素が入っているとき
            {
                Node temp = stk.Pop();//スタックの先頭を出し、スタックから削除
                foreach(int x in temp.child)
                {
                    nodes[x].depth = temp.depth + 1;
                    if(nodes[x].child.Count>0)//leafに行きつくまで繰り返し
                    {
                        stk.Push(nodes[x]);
                    }
                }
            }
            for(int i=0; i<n; i++)
            {
                StringBuilder Array = new StringBuilder();//名前空間はSystem.text
                int a= nodes[i].child.Count;//nodeによって変わるchid配列の長さを頑張って出力
                if (a >= 1)
                {
                    Array.Append(nodes[i].child[0].ToString());
                }
                if (a >= 2)
                {
                    for(int j=1; j<a; j++)
                    {
                        Array.Append(", ");
                        Array.Append(nodes[i].child[j].ToString());
                    }
                }
                Console.WriteLine("node {0}: parent = {1}, depth = {2}, {3}, [{4}]",i,nodes[i].parent,nodes[i].depth,nodes[i].type,Array);
            }
        }

    }
}