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++14 (judge verdict: Accepted)

#include<bits/stdc++.h>

#define REP(i,n) for(int (i)=0;(i)<(n);++(i))
#define FOR(i,a,b) for(int (i)=(a);i<(b);++(i))
#define FOR_DOWN(i,a,b) for(int (i)=(a);i>=(b);--(i))
#define ALL(obj) (obj).begin(),(obj).end()
#define space <<" "<<
#define maxs(x,y) (x=max(x,y))
#define mins(x,y) (x=min(x,y))
#define FIX fixed<<setprecision(10)
using namespace std;

using ll = long long; using ull = unsigned long long;
using vb = vector<bool>; using vvb = vector<vb>;
using vi = vector<int>; using vvi = vector<vi>;
using vl = vector<ll>; using vvl = vector<vl>;
using pii = pair<int,int>; using pll = pair<ll,ll>;

const ll MOD = 1e9+7;

class Stack{
private:
char m[20000];
int tp;
public:
    Stack(){
    tp=0;
    }
    void push(char x){
        tp++;
        m[tp]=x;
    }
    char pop(){
        tp--;
        return m[tp+1];
    }
}; 
class Queue{
private:
pair<string,int> m[60000];
int head,tail;
public:
    Queue(){
        head=0;
        tail=0;
    }
    bool isEmpty(){
        return head==tail;
    }
    void enqueue(pair<string,int> x){
        m[tail]=x;
        if(tail +1 ==60000)tail=0;
        else tail++;
    }
    pair<string,int> deque(){
        pair<string,int> x=m[head];
        if(head+1==60000)head=0;
        else head++;
        return x;
    } 
};

struct Node{
    int key;
    Node* prev;
    Node* next;
};

class LinkedList{
public:
LinkedList(){
    dammy= new Node();
    dammy->key =0;
    dammy->prev= dammy;
    dammy->next= dammy;
}
~LinkedList(){
    Node* tmpNode =dammy->next;
    Node* nextNode;
    while(tmpNode != dammy){
        nextNode=tmpNode->next;
        delete tmpNode;
        tmpNode=nextNode;
    }
    delete dammy;
}
void insertNode(int x){
    Node* currentNode=dammy->next;
    dammy->next=new Node();
    dammy->next-> key =x;
    dammy->next->next=currentNode;
    dammy->next->prev=dammy;
    currentNode->prev = dammy->next;
}
    Node* find(int key){
    for(Node* i=dammy->next;i!=dammy;i=i->next){
      if(i->key==key)return i;
    }
    return dammy;
    }
    void deleteNode(int x){
    Node* currentNode = find(x);
    if(currentNode==dammy)return;
    currentNode->prev->next=currentNode->next;
    currentNode->next->prev=currentNode->prev;
    delete currentNode;
    }
    void deleteFirst(){
        Node* currentNode = dammy->next;
        dammy->next = dammy->next->next;
        dammy->next->prev = dammy;
        delete currentNode; 
    };
 
    void deleteLast(){
        Node* currentNode = dammy->prev;
        dammy->prev = dammy->prev->prev;
        dammy->prev->next = dammy;
        delete currentNode; 
    };

    void printList(){
    Node* currentNode = dammy->next;;
    int sp = 1;
    while(1){
    if ( currentNode == dammy ) break;
    if ( sp == 0)  cout<<" ";
    cout<<currentNode->key;
    currentNode = currentNode->next;
    sp = 0;
    }
    cout<<endl;
    }
    private:
    Node* dammy;
};
    int n;
void CountingSort(int* A,int* B,int k){
    int C[k+1];
    REP(i,k+1){
        C[i]=0;
    }
    REP(i,n){
        C[A[i+1]]++;
    }
    FOR(i,1,k+1){
        C[i]+=C[i-1];
    }
    for(int i=n;i>=1;i--){
        B[C[A[i]]]=A[i];
        C[A[i]]--;
    }
}

struct tramp{
    char mark;
    int number;
    int order;
};

void sw(tramp *a,tramp* b){
    tramp temp = *a;
    *a=*b;
    *b=temp;
}
int partition(tramp* A,int p, int r){
    int x= A[r].number;
    int i=p-1;
    for(int j=p;j<r;j++){
        if(A[j].number<=x){
            i=i+1;
            sw(&A[i],&A[j]);
        }
    }
    sw(&A[i+1],&A[r]);
    return i+1;
}
void quick(tramp*A,int p,int r){
    if(p<r){
        int q=partition(A,p,r);
        quick(A,p,q-1);
        quick(A,q+1,r);
    }
}
struct point{
    int parent;
    int depth;
    int childNumber;
    vector<int> children;
    point(){
        parent=-1;
        depth=0;
    }
};

int main(){
    int n;
    cin>>n;
    point points[n];
    for(int i=0;i<n;i++){
        int id;
        cin>>id;
        cin>>points[id].childNumber;
        for(int j=0;j<points[id].childNumber;j++){
            int k;
            cin>>k;
            points[id].children.push_back(k);
            points[k].parent=id;
        }
    }

    for(int i=0;i<n;i++){
        int j=points[i].parent;
        int temp;
        temp=j;
        while (j!=-1)
        {
            points[i].depth+=1;
            j=points[temp].parent;
            temp=j;
        }
    }

    for(int i=0;i<n;i++){
        cout<<"node "<<i<<": parent = "<<points[i].parent<<", depth = "<<points[i].depth<<", "<<
        (points[i].parent==-1? "root":(points[i].childNumber==0?"leaf" :"internal node"))<<", [";
        if(points[i].childNumber>0){
        for(int j=0;j<points[i].childNumber-1;j++){
            cout<<points[i].children.at(j)<<", ";
        }
        cout<<points[i].children.at(points[i].childNumber-1);
        }
        cout<<"]"<<endl;
    }
}