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)

#include<stdio.h>
#define MAX 100000

int confirm_depth(int);

struct Node// This variable is used only confirm the node is root,internal,or leaf.
{
  int parent , left , right;
};

/*
struct insert //This variable is used to remember how many children the node has.
{
  int id,number_of_children;
};
*/

//struct insert in[100000];
struct Node node[MAX] = {-1,-1,-1};

int main()
{
  
  int n,k,i,j,id,child_id,next,count=0;
  scanf("%d",&n);
  for(i = 0 ; i<n ;i++)
    {
      node[i].parent = -1;
      node[i].left = -1;
      node[i].right = -1;
      //printf("node[0]={%d,%d,%d}\n",node[0].parent,node[0].left,node[0].right);
    }
     
  for(i = 0 ;i < n ; i++ )
    {
      //scanf("%d %d",&in[i].id,&in[i].number_of_children);
      scanf("%d %d",&id,&k);
      //for(j =count ; j < count+ in[i].number_of_children ; j++)
      //printf("TEST node[%d].left = %d\n",i,node[i].left);
      for(j =count ; j < count+k ; j++)
	{

	  next = child_id;
	  scanf("%d",&child_id);
	  
	  //if(j>count)node[id].right = child_id;
	  //printf("node[%d].right = %d\n",id,node[id].right);
	  if(j == count)
	    {
	      node[id].left = child_id;
	      //printf("j = %d\n",j);
	      //printf("id = %d\n",id);
	      //printf("node[%d].left = %d\n",id,node[id].left);
	    }
	  
	  if(j>count )
	    {
	      if(j == count +1)
		{
		  
		  node[node[id].left].right = child_id;
		  
		  //printf("\nnode[%d].right = %d\n\n",id,node[id].right);
		}
	      else node[next].right = child_id;
	      //printf("child_id = %d\n",child_id);
	      //printf("node[%d].right = %d\n",next,node[next].right);
	    }
	  //node[child_id].parent  = in[i].id;
	  node[child_id].parent  = id;
	  
	  
	    
	}
      count = count + k;
      //  printf("count = %d\n",count);
      
    }

  //printf("TEST node[%d].left = %d\n",6,node[6].left);



  
  //printf("\n\n");
  for(i = 0; i <n ; i++)
    {
      //printf("TEST node[%d].left = %d\n",i,node[i].left);
      printf("node %d: parent = %d, depth = %d, ",i,node[i].parent,confirm_depth(i));
      if(node[i].parent == -1) printf("root, ");
      else if(node[i].left != -1) printf("internal node, ");
      else printf("leaf, ");

      printf("[");
      if(node[i].left >= 0)
	{

	  next = node[i].left;
	  //printf("  node[%d].left = %d\n",i,node[i].left);
	  //printf("[");
	   if(node[i].left >= 0)printf("%d",node[i].left);
	  
	  while(node[next].right >=0)
	    {
	      
	      printf(", %d",node[next].right);
	      next = node[next].right;
	    }
	  
	 
	}
      printf("]");
      printf("\n");
    }

}
  

int confirm_depth(int id)
{
  int i,depth = 0;
  while(node[id].parent != -1 )
    {
      id = node[id].parent;
      depth++;
    }
  return depth;
}
  
  /*
  printf("\n\n");
  //printf("%d\n",NULL);
  for(i = 0; i <n ; i++)
    {
      printf("node %d:parent = %d property = %d \n ",i,node[i].parent,node[i].right);
      printf("node[%d].left = %d\n",i,node[i].left);
      
      if(node[i].left >= 0)
	{
	  printf("test\n");
	  next = node[i].left;
	  printf("node[%d].left = %d\n",i,node[i].left);
	  
	  while(next >=0)
	    {
	     
	      printf("right = %d\n",node[next].right);
	      printf("next = %d\n ",next);
	      next = node[next].right;
	      printf("next[next.right] = %d\n ",next);
	      break;
	      
	    }
	  printf("\n");
	}
    }

}
  */