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>
#include<stdlib.h>

#define MAX 100000
#define ROOT 0
#define INODE 1
#define LEAF 2

struct CHILDREN {
	int id;
	struct CHILDREN *next;
};
typedef struct CHILDREN CHILDREN;

struct NODE {
	int parent;
	short depth;
	short type;
	CHILDREN *children;
};
typedef struct NODE NODE;

int main( void ) 
{   
    int n;
    int id,k,c,prec;
    int i,j,depth;
    char types[3][32] = { "root", "internal node", "leaf" };
	NODE nodes[ MAX+1 ];
	CHILDREN *child=NULL,*pre=NULL;
	       
    scanf( "%d", &n );
    for( i=0; i<n; i++ ) {
        scanf( "%d%d", &id, &k );
        if( k ) {
       		prec=id;
       		child = (CHILDREN *)malloc( sizeof( CHILDREN ) );
       		nodes[id].children=child;
       		pre=child;
        	for( j=0; j<k; j++ ) {
	           	scanf( "%d", &c );
	           	if( !nodes[c].type ) {
          			nodes[c].type=INODE;        	
        		}
	           	nodes[c].parent=id;
	           	nodes[c].depth=nodes[id].depth+1;
	           	child->id=c;
	           	pre->next=child;
	           	pre=child;
	           	if( j < k-1 ) {
	      			child = (CHILDREN *)malloc( sizeof( CHILDREN ) );
	      		}
        	}
        	if( 0 == nodes[id].depth ) {
        		nodes[id].parent = -1; 
        		nodes[id].type=ROOT;
        	}        	
        	pre->next=NULL;     	
       	} else {
       		nodes[id].type=LEAF;
	   		nodes[id].children=NULL;
       	}
    }
    if( 1 == n ) {
    	nodes[id].parent = -1; 
    	nodes[id].type=ROOT;
   	}
    
    for( i=0; i<n; i++ ) {
    	depth=(-1 == nodes[i].parent)? 0: nodes[ nodes[i].parent ].depth+1;
		printf( "node %d: parent = %d, depth = %d, %s, ", i, nodes[i].parent,  depth,  types[ nodes[i].type ] );
		if( NULL != nodes[i].children ) {
			child=nodes[i].children;
			printf( "[%d", child->id );
			child=child->next;
			while( NULL != child ) {
				printf( ", %d", child->id );
				child=child->next;
			}
			printf( "]\n" );
		} else {
			printf( "[]\n" );
		}
	}

    return 0;
}