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

typedef struct{
    int nodeID;
    int childNum;
    int *child;
    int parent;
    int depth;
}Node;

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

void selectionSort(int,Node[]);
int binSearch(int,int[],Node);

int main(int argc, char** argv) {
    int size, childNum, id;
    scanf("%d\n", &size);
    Node tree[size];
    for(int i = 0; i < size; i++){
        scanf("%d %d", &id, &childNum);
        initNode(&tree[i],id,childNum);
        for(int j = 0; j < childNum; j++)
            scanf("%d", &tree[i].child[j]);
    }
    selectionSort(size,tree);
    for(int i = 0; i < size; i++){
        tree[i].parent = getParent(size,tree,tree[i]);
        tree[i].depth = -1;
    }
    printTree(size, tree);
    return (EXIT_SUCCESS);
}

void initNode(Node *n, int id, int num){
    int *temp = (int*)malloc(num * sizeof(int));
    n->nodeID = id;
    n->childNum = num;
    n->child = temp;
}

int getParent(int size, Node tree[size], Node target){
    for(int i = 0; i < size; i++)
        for(int j = 0; j < tree[i].childNum; j++)
            if(target.nodeID == tree[i].child[j])
                return tree[i].nodeID;
    return -1;
}

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

int getDepth(int size, Node tree[size], Node target){
    int count = 0;
    if(target.parent >= 0 && tree[target.parent].depth == -1){
        count += getDepth(size,tree,tree[target.parent]);
        count++;
    }
    else if(tree[target.parent].depth >= 0)
        count = tree[target.parent].depth;
    return count;
}

int getNodeType(int size, Node tree[size], Node target){    //Return types: -1 = Root
    if(target.parent == -1)
        return -1;
    else{
        if(target.childNum == 0)
            return 1;
        return 0;
    }
}

void selectionSort(int size, Node array[size]){
    int minLoc;
    Node temp;
    for(int i = 0; i < size; i++){
        minLoc = i;
        for(int j = i; j < size; j++)
            if(array[j].nodeID < array[minLoc].nodeID)
                minLoc = j;
        temp = array[i];
        array[i] = array[minLoc];
        array[minLoc] = temp;
    }
}

int binSearch(int size, int array[size], Node searchVal){
    int left = 0;
    int right = size;
    int mid;
    while(left < right){
        mid = (left + right)/2;
        if(array[mid] == searchVal.nodeID)
            return mid;
        if(searchVal.nodeID > array[mid])
            left = mid + 1;
        else if(searchVal.nodeID < array[mid])
            right = mid;
    }
    return -1;
}