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>
typedef struct Node Node;
struct Node{
    int id;
    Node* parent;
    Node* rcbling;
    Node* lchild;
};

void initialize(Node* n);
int findDepth(Node* a);
int main(int argc, char** argv) {
    int size;
    scanf("%d", &size);
    Node array[size];
    
    int degree;
    int location, tempP, tempC = -1;
    for(int i=0;i<size;i++)
    {
        initialize(&array[i]);
    }
    
    for(int i=0;i<size;i++)
    { 
        scanf("%d %d", &tempP, &degree);
        array[tempP].id = tempP;
        for(int j=0;j<degree;j++)
        {
           
            scanf("%d",&location);
            array[location].parent=array + tempP;
            if(j==0)
            {
                array[tempP].lchild = array + location;
            }
            if(tempC>=0)
            {
                array[tempC].rcbling = array + location;
            }
            tempC = location;
        }
        tempC = -1;
    }
    
    int temp, temp2P, type;
    for(int i=0; i<size; i++)
    {
        printf("node %d: ", i);
        if(array[i].parent == NULL)
            temp2P = -1;
        else
            temp2P = array[i].parent->id;
        printf("parent = %d, ", temp2P);
        printf("depth = %d, ", findDepth(&array[i]));
        if(array[i].parent==NULL)
            type = 1;
        else{
            if(array[i].lchild!=NULL)
                type = 2;
            else
                type = 3;
        }
        if(type == 1)
        {
            printf("root, [");
        }
        if(type == 2)
        {
            printf("internal node, [");
        }
        if(type == 3)
        {
            printf("leaf, [");
        }
        if(array[i].lchild != NULL){
            temp = array[i].lchild->id;
            printf("%d", temp);
            while(array[temp].rcbling!=NULL){
                if(array[temp].rcbling != NULL)
                    printf(", ");
            temp = array[temp].rcbling->id;
            printf("%d", temp);
            }
        }
        printf("]\n");
        
    }
    
    return (EXIT_SUCCESS);
}


void initialize(Node* n)
{
    n->id = -1;
    n->lchild = NULL;
    n->rcbling = NULL;
    n->parent = NULL;
}

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