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 <string.h>
#include <stdlib.h>

struct Node {
	int key;
	struct Node* parent;
	int degree;
	int depth;
	int height;
	char type[15];
	
	int * child_list;
};
typedef struct Node * NodePointer;


void serach_depth (NodePointer ,int  ,NodePointer[]);

int main (){
	int n;
	scanf("%d\n",&n);
	
	//setting part
	NodePointer array [n];
	for (int i = 0; i < n;i++)
	{
		array[i] = (NodePointer) malloc (sizeof(struct Node));
		array[i]->parent = NULL;
		array[i]->degree = 0;
		strcpy(array[i]->type,"internal node");
	}
	
	//input part
	for (int i = 0; i < n;i++)
	{
		int id;
		scanf("%d ",&id);
		array[id]->key = id;
		int child_num;
		scanf("%d" ,&child_num);
		array[id]->child_list = (int *) malloc (child_num * sizeof(int));
		
		if ( child_num == 0){
			strcpy(array[id]->type ,"leaf");
		}else {
			for (int j = 0; j < child_num; j++){
				int child_id;
				scanf(" %d" ,&child_id);
				array[id]->child_list[j] = child_id;
				array[child_id]->parent = array[id];
				array[id]->degree++;
			}
		}

	}
	
	//search part
		for (int i = 0; i < n;i++)
	{
		if(array[i]->parent == NULL) {
			strcpy(array[i]->type ,"root");
			serach_depth(array[i],0 ,array); break;
		}
	}
	
	//print part
	for (int i = 0; i < n ;i++){
		printf("node %d: ",i);
		if (array[i] -> parent != NULL){
		printf("parent = %d, ",array[i]->parent->key);
		}else {
			printf("parent = -1, ");
		}
		printf("depth = %d, %s, [",array[i]->depth,array[i]->type );
		for ( int j = 0; j < array[i]->degree; j++){
			printf("%d" ,array[i]->child_list[j]);
			if ( j + 1 != array[i]->degree){printf(", ");}
		}
		printf("]\n");
	}
	
	//printf("%d" ,array[6]->child_list[3]);
	
	return 0;
}

void serach_depth (NodePointer tag,int dep , NodePointer array[]){
	tag->depth = dep;
	if ( tag->degree == 0 ){tag->height = 0; return;}
	for ( int j = 0; j < tag->degree ; j++){
		serach_depth(array [ (tag->child_list[j]) ] , dep + 1 ,array);
	}
}