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)

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/* 
 * File:   main.c
 * Author: Mu-Chan
 *
 * Created on July 9, 2018, 6:18 PM
 */

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

typedef struct Node Node;

struct Node{
    int id;
    Node* parent;
    Node* sibling;
    Node* child;
};

typedef struct{
    Node *array;
}Tree;

void initNode(Node*);
void printTree(int,Node[]);
int getDepth(Node*);

int main(int argc, char** argv) {
    int size;
    scanf("%d\n", &size);
//    Tree tree;
    Node array[size];
    int loc, childNum, tempParent;
    int tempChild = -1;
    
    for(int i = 0; i < size; i++)
        initNode(&array[i]);
    
    for(int i = 0; i < size; i++){
        scanf("%d %d", &tempParent, &childNum);
        array[tempParent].id = tempParent;
        for(int j = 0; j < childNum; j++){
            scanf("%d", &loc);
            array[loc].parent = array+tempParent;
            if(j == 0)
                array[tempParent].child = array+loc;
            if(tempChild >= 0)
                array[tempChild].sibling = array+loc;
            tempChild = loc;
        }
        tempChild = -1;
    }
//    tree.array = array;
    printTree(size,array);
    
    return (EXIT_SUCCESS);
}

void initNode(Node *n){
    n->id = -1;
    n->child = NULL;
    n->sibling = NULL;
    n->parent = NULL;
}

void printTree(int size, Node tree[size]){
    int parent;
    char type[15];
    int temp;
    for(int i = 0; i < size; i++){
        printf("node %d: ", i);
        if(tree[i].parent == NULL)
            parent = -1;
        else
            parent = tree[i].parent->id;
        printf("parent = %d, ", parent);
        printf("depth = %d, ", getDepth(&tree[i]));
        if(tree[i].parent == NULL)
            strcpy(type,"root");
        else{
            if(tree[i].child != NULL)
                strcpy(type,"internal node");
            else
                strcpy(type,"leaf");
        }
        printf("%s, [", type);
        if(tree[i].child != NULL){
            temp = tree[i].child->id;
            printf("%d", temp);
            while(tree[temp].sibling != NULL){
                if(tree[temp].sibling != NULL)
                    printf(", ");
                temp = tree[temp].sibling->id;
                printf("%d", temp);
            }
        }
        printf("]\n");
    }
}

int getDepth(Node* target){
    int count = 0;
    if(target->parent != NULL){
        count++;
        count += getDepth(target->parent);
    }
    return count;
}