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 JAVA (judge verdict: Accepted)

import java.util.Scanner;

class Node {
   private int parent;
   private int left;
   private int right;

   public Node(int parent, int left, int right) {
       this.parent = parent;
       this.left = left;
       this.right = right;
   }
   public void setLeft(int left) {
       this.left = left;
   }
   public void setRight(int right) {
       this.right = right;
   }
   public void setParent(int parent) {
       this.parent = parent;
   }
   public int getParent() {
       return this.parent;
   }
   public int getLeft() {
       return this.left;
   }
   public int getRight() {
       return this.right;
   }
}
public class Main {
    static final int NIL = -1;
    static final int MAX = 100005;
    static Node[] T = new Node[MAX];
    static int n;
    static int D[] = new int[MAX];

    public static void main(String[] args){
        int left = 0;

        Scanner scanner = new Scanner(System.in);
        n = scanner.nextInt();

        for (int i = 0; i < n; i++) {
            T[i] = new Node(NIL, NIL, NIL); // 初始化节点
        }
        for (int i = 0; i < n; i++) {
            int number = scanner.nextInt();
            int childrenSize = scanner.nextInt();
            for (int j = 0; j < childrenSize; j++) {
                int child = scanner.nextInt();
                // 左子右兄弟
                if (j == 0) {
                    T[number].setLeft(child);
                } else {
                    T[left].setRight(child);
                }
                left = child;
                T[child].setParent(number);
            }
        }
        scanner.close();
        // 求各节点的深度
        for ( int i = 0; i < n; i++) {
            D[i] = getDepth(i);
        }

        for (int i = 0; i < n; i++) {
            print(i);
        }
    }

    static int getDepth(int number) {
        int d = 0;
        while(T[number].getParent() != NIL) {
            number = T[number].getParent();
            d ++;
        }
        return d;
    }

    static void print(int number) {
        System.out.print("node " + number + ": " + "parent = " + T[number].getParent() + ", " + "depth = " + D[number] + ", ");

        if (T[number].getParent() == NIL) {
            System.out.print("root, ");
        } else if (T[number].getLeft() == NIL) {
            System.out.print("leaf, ");
        } else {
            System.out.print("internal node, ");
        }

        System.out.print("[");

        for (int i = 0, c = T[number].getLeft(); c != NIL; i++, c = T[c].getRight()){
            if(i != 0) {
                System.out.print(", ");
            }
            System.out.print(c);
        }
        System.out.println("]");
    }
}