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 SIZE 100001

enum TYPE{root, node, leaf, null};
char* strtype[] = {"root", "internal node", "leaf"};

typedef struct Vec{
	int lenght;
	int size;
	int maxsize;
	int* buff;
}VEC;

typedef struct VecIt{
	int i;
	VEC* vp;
}VECIT;

struct TREE{
	int parent;
    int depth;
	int type;
	VEC* vp;
};

struct TREE tree[SIZE];
int treeroot;

VEC* newVec(void)
{
	VEC* vp = (VEC*)malloc(sizeof(VEC));
	vp->lenght = 0;
	vp->size = 10;
	vp->maxsize = SIZE;
	vp->buff = (int*)calloc(vp->size, sizeof(int));
	return(vp);
}

void delVec(VEC* vp)
{
	free(vp->buff);
	free(vp);
	return;
}

void resizeVec(VEC* vp)
{
	vp->size += vp->size;
	vp->buff = (int*)realloc(vp->buff, vp->size*sizeof(int));
	return;
}

void addVec(VEC* vp, int n)
{
	if(vp->size <= vp->lenght){
		resizeVec(vp);
	}
	vp->buff[vp->lenght] = n;
	vp->lenght++;
	return;
}

int getVec(VEC* vp, int n)
{
	if(n < vp->lenght){
		return(vp->buff[n]);
	}
	return(-1);
}

int lenghtVec(VEC* vp)
{
	return(vp->lenght);
}

VECIT* newVecIt(VEC* vp)
{
	VECIT* vpit = (VECIT*)malloc(sizeof(VECIT));
	vpit->vp = vp;
	vpit->i = 0;
	return(vpit);
}

void delVecIt(VECIT* vp)
{
	free(vp);
	return;
}

int nextVecIt(VECIT* vp)
{
	return(getVec(vp->vp, vp->i++));
}

void init(void)
{
	int i;
	for(i = 0; i < SIZE; i++){
		tree[i].parent = -1;
		tree[i].depth = 0;
		tree[i].type = null;
		tree[i].vp = newVec();
	}
	return;
}

void end(void)
{
	int i;
	for(i = 0; i < SIZE; i++){
		delVec(tree[i].vp);
	}
	return;
}

void tr(int node, int lv)
{
	int n;
	VECIT* it = newVecIt(tree[node].vp);
	while((n = nextVecIt(it)) != -1){
		tr(n, lv + 1);
	}
	tree[node].depth = lv;
	return;
}

int main(int argc, char* argv[])
{
	int i, j, k, n, c, id, itn;
	VECIT* it;

	init();

	scanf("%d", &n);
	for(i = 0; i < n; i++){
		scanf("%d", &id);
		scanf("%d", &k);
		for(j = 0; j < k; j++){
			scanf("%d", &c);
			if(tree[c].type == null){
			  tree[c].type = leaf;
			}
			tree[c].parent = id;
			addVec(tree[id].vp, c);
		}
		if(0 < lenghtVec(tree[id].vp)){
			tree[id].type = node;
		}else{
			tree[id].type = leaf;
		}
	}

	for(i = 0; i < n; i++){
		if(tree[i].parent == -1){
			treeroot = i;
			break;
		}
	}
	tree[treeroot].type = root;
	tr(treeroot, 0);

	for(i = 0; i < n; i++){
		printf("node %d: parent = %d, depth = %d, ", i, tree[i].parent, tree[i].depth);
		printf("%s, [", strtype[tree[i].type]);
		it = newVecIt(tree[i].vp);
		itn = lenghtVec(tree[i].vp);
		for(j = 0; j < itn-1; j++){
			printf("%d, ", nextVecIt(it));
		}
		if(itn == 0){
			printf("]\n");
		}else{
		    printf("%d]\n", nextVecIt(it));
		}
		
		delVecIt(it);
	}
	
	end();

	return(0);
}