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: Wrong Answer)

using System;
using System.Text;

namespace ALDS1_7_A
{
    public struct node
    {
        public int parent;
        public int left;
        public int right;

        public node(int parent, int left, int right)
        {
            this.parent = parent;
            this.left = left;
            this.right = right;
        }
    }

    class Program
    {
        public static int nil = -1;

        static int getdepth(int u, node[] T)
        {
            int d = 0;
            while (T[u].parent != nil)
            {
                u = T[u].parent;
                d++;
            }
            return d;
        }

        static int[] D = new int[100000];

        static void setdepth(int u, int p,node[] T)
        {
            D[u] = p;
            if (T[u].right != nil)
            {
                setdepth(T[u].right, p,T);
            }
            if (T[u].left != nil)
            {
                setdepth(T[u].left, p + 1,T);
            }
        }
        
        static StringBuilder sbchildren(int u,node[] T)
        {
            var sb = new StringBuilder();
            sb.Append(" [");
            int c = T[u].left;

            while (c !=nil)
            {
                sb.Append(c+", ");
                c = T[c].right;
            }

            sb.Remove(sb.Length-2, 2);

            return sb;
        }

        static int rootchecker(node[] T)
        {
            for (int i = 0; i < T.Length; i++)
            {
                if (T[i].parent == nil)
                {
                    return i;
                }
            }
            return 0;
        }
 
        static void Main(string[] args)
        {
            int n = getint();

            node[] T = new node[n];
            for (int i = 0; i < n; i++)
            {
                T[i].parent = nil;
                T[i].left = nil;
                T[i].right = nil;
            }
            
            for (int i = 0; i < n; i++)
            {
                var input = getsplit();
                int[] a = Array.ConvertAll(input, int.Parse);
                if (a[1] >= 1)
                {
                    T[a[0]].left = a[2];
                    int j;
                    for (j = 2; j <= a[1]; j++)
                    {
                        T[a[j]].right = a[j + 1];
                        T[a[j]].parent = a[0];
                    }
                   
                    T[a[j]].parent = a[0];
                }
              
            }

            int r = rootchecker(T);

            setdepth(r, 0, T);

            for (int i = 0; i < n; i++)
            {
                var sb = new StringBuilder();
                sb.Append("node " + i + ": ");
                sb.Append("parent = " + T[i].parent + ", ");
                sb.Append("depth = " + D[i] + ", ");

                if (D[i] == 0)
                {
                    sb.Append("root,");
                    sb.Append(sbchildren(i, T));
                    sb.Append("]");
                }
                else
                {
                    
                    if (sbchildren(i,T).Length == 0)
                    {
                        sb.Append("leaf, []");
                    }
                    else
                    {
                        sb.Append("internal node,");
                        sb.Append(sbchildren(i, T));
                        sb.Append("]");
                    }
                }
                Console.WriteLine(sb);
            }          
        }

        public static string[] getsplit()
        {
            string[] x = Console.ReadLine().Split(' ');
            return x;
        }
        public static int getint()
        {
            int x = int.Parse(Console.ReadLine());
            return x;
        }
    }
}