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: Time Limit Exceeded)

#include <stdio.h>
#define N 100000

typedef struct node {
  int id;
  int depth;
  struct node* parent;
  struct node* child;
  struct node* next;
} NODE;

void append(int, NODE*, NODE*);
void calcDepth(NODE*);

int main(){
  int i, j, n;
  int root;
  int id, nchild;
  int childid;
  NODE* last;
  NODE* tmp;
  NODE node[N];

  // input the number of nodes
  scanf("%d", &n);

  // input the information of nodes
  for(i = 0; i < n; i++) {
    scanf("%d", &id);  // input the node id
    scanf("%d", &nchild);  // input the number of children
    for(j = 0; j < nchild; j++) {
      scanf("%d", &childid);  // input the child id
      node[id].id = id;
      append(childid, &node[id], &node[childid]);
    }
  }

  //depth
  i = 0;
  while(1) {
    if(node[i].parent == NULL){
      root = i;
      break;
    }
    i++;
  }
  
  for(i = 0; i < n; i++) {
    calcDepth(&node[root]);
  }
  
  // output
  for(i = 0; i < n; i++) {
    // node id
    printf("node %d: ", i);
    // the parent of node
    if(node[i].parent == NULL) printf("parent = -1, ");
    else printf("parent = %d, ", node[i].parent->id);
    // node depth
    printf("depth = %d, ", node[i].depth);
    // the type of node
    if(node[i].parent == NULL) printf("root, ");
    else if(node[i].child == NULL) printf("leaf, ");
    else printf("internal node, ");
    // child
    last = node[i].child;
    printf("[");
    while(last != NULL) {
      printf("%d", last->id);
      last = last->next;
      if(last != NULL) printf(", ");
    }
     
    
    printf("]\n");
  }

  return 0;
}

// function: add to the tree
void append(int id, NODE* parent, NODE* node) {
  NODE* last;

  node->id = id;
  node->parent = parent;
  node->depth = 0;
  node->next = NULL;
  if(parent == NULL){
    return;
  }
  if(parent->child == NULL) {
    parent->child = node;
  }
  else {
    last = parent->child;
    while(last->next != NULL) {
      last = last->next;
    }
    last->next = node;
  }
}

void calcDepth(NODE* node) {
  NODE* last;
  NODE* tmp;
  
  last = node;
  if(last->parent == NULL){
    last->depth = 0;
  }
  else if(last->child == NULL) return;
  
  tmp = last;
  last = last->child;
  while(last != NULL) {
    last->depth = tmp->depth + 1;
    calcDepth(last);
    last = last->next;
  }
}