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 Scala (judge verdict: Wrong Answer)


import scala.collection.mutable.ArrayBuffer
import scala.io.StdIn

object Main extends App {

  final val nil = -1
  
  case class Node(var parent: Int = -1,var left: Int = -1, var right: Int = -1){
    
    def typeInfo: String = if(parent == -1) {
      "root"
    } else if(left == nil ) {
      "leaf"
    } else {
      "internal node"
    }
    
    def nodeLiteral(depth: Int, childs: Array[Int]): String = {
      s"parent = $parent, depth = $depth, $typeInfo, [${childs.mkString(", ")}]"
    }
    override def toString: String = s"parent = $parent, depth = X, $typeInfo, [child]"
    
  }
  
  def setDepth(index: Int, cnt: Int): Unit = {
    depth(index) = cnt
    if(input(index).right != nil){ // brother
      setDepth(input(index).right, cnt)
    }
    if(input(index).left != nil){ // child
      setDepth(input(index).left, cnt + 1)
    }
  }
  
  def getChild(index: Int): Array[Int] = {
    if(input(index).left == nil){
      Array.empty[Int]
    } else {
      var node = input(input(index).left)
      var ret = ArrayBuffer(input(index).left)
      while(node.right != nil){
        ret += node.right
        node = input(node.right)
      }
      ret.toArray
    }
  }
  
  val n = StdIn.readLine().trim.toInt
  val input = Array.fill[Node](n)(Node())
  val depth = Array.fill[Int](n)(0)

  (0 until n).foreach{ _ =>
    val a = StdIn.readLine().trim.split(" ").map(_.toInt)
    val i = a(0)
    val c = a.drop(2)
    var before: Int = -1
    c.zipWithIndex.foreach{ ci =>
      val childNode = Node(i, nil, nil)
      input(ci._1) = childNode
      if(ci._2 == 0) input(i).left =  ci._1
      if(before != nil){
        input(before).right = ci._1
      }
      before = ci._1
    }
  }

  var root = nil
  (0 until n).foreach{ i =>
    if(input(i).parent == nil) root = i
  }
  
  setDepth(root, 0)
  
  (0 until n).foreach{ i =>
    val ch = getChild(i)
    println(s"node $i: " + input(i).nodeLiteral(depth(i), ch))
  }

}