Problem:
Binary Tree

A rooted binary tree is a tree with a root node in which every node has at most two children.

Your task is to write a program which reads a rooted binary tree T and prints the following information for each node u of T:

node ID of u

parent of u

sibling of u

the number of children of u

depth of u

height of u

node type (root, internal node or leaf)

If two nodes have the same parent, they are siblings. Here, if u and v have the same parent, we say u is a sibling of v (vice versa).

The height of a node in a tree is the number of edges on the longest simple downward path from the node to a leaf.

Here, the given binary tree consists of n nodes and evey node has a unique ID from 0 to n-1.

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 is given in the following format:

id left right

id is the node ID, left is ID of the left child and right is ID of the right child. If the node does not have the left (right) child, the left(right) is indicated by -1.

Output

Print the information of each node in the following format:

node id: parent = p, sibling = s, degree = deg, depth = dep, height = h, type

p is ID of its parent. If the node does not have a parent, print -1.

s is ID of its sibling. If the node does not have a sibling, print -1.

deg, dep and h are the number of children, depth and height of the node respectively.

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.

Please follow the format presented in a sample output below.

Constraints

1 ≤ n ≤ 25

Sample Input 1

9
0 1 4
1 2 3
2 -1 -1
3 -1 -1
4 5 8
5 6 7
6 -1 -1
7 -1 -1
8 -1 -1

Sample Output 1

node 0: parent = -1, sibling = -1, degree = 2, depth = 0, height = 3, root
node 1: parent = 0, sibling = 4, degree = 2, depth = 1, height = 1, internal node
node 2: parent = 1, sibling = 3, degree = 0, depth = 2, height = 0, leaf
node 3: parent = 1, sibling = 2, degree = 0, depth = 2, height = 0, leaf
node 4: parent = 0, sibling = 1, degree = 2, depth = 1, height = 2, internal node
node 5: parent = 4, sibling = 8, degree = 2, depth = 2, height = 1, internal node
node 6: parent = 5, sibling = 7, degree = 0, depth = 3, height = 0, leaf
node 7: parent = 5, sibling = 6, degree = 0, depth = 3, height = 0, leaf
node 8: parent = 4, sibling = 5, degree = 0, depth = 2, height = 0, leaf

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Ruby (judge verdict: Wrong Answer)

# ALDS1_7_B: Binary Tree

class Node
  class << self
    def bind(parent, left, right)
      parent.set_children(left, right)
      left.set_parent(parent, :left)
      right.set_parent(parent, :right)
    end
  end

  attr_accessor :id, :left, :right
  attr_reader :depth, :height

  def initialize(id)
    @id = id
    @depth = @height = 0
  end

  def update_tree
    update_depths
    update_heights
  end

  def parent
    root? ? NIL : @parent
  end

  def sibling
    return NIL if root?
    on_left? ? @parent.right : @parent.left
  end

  def children
    [@left, @right].select { |n| not n.nil_node? }
  end

  def degree
    children.size
  end

  def type
    case
    when root? then 'root'
    when degree == 0 then 'leaf'
    else 'internal node'
    end
  end

  def root?
    @parent.nil?
  end

  def nil_node?
    self == NIL
  end

  def on_left?
    @position == :left
  end

  def set_children(left, right)
    @left  = left
    @right = right
  end

  def set_parent(parent, position)
    @parent   = parent
    @position = position
  end

  protected
    attr_writer :depth, :height

    def update_depths(depth = 0)
      return if nil_node?
      @depth = depth
      @left.update_depths @depth + 1
      @right.update_depths @depth + 1
    end

    def update_heights
      return -1 if nil_node?
      lh = @left.update_heights
      rh = @right.update_heights
      @height = (lh < rh ? lh : rh) + 1
    end
end

Node.const_set :NIL, Node.new(-1)

n_nodes = gets.to_i
tree = Array.new(n_nodes).map.with_index do |_, i|
  Node.new(i)
end

n_nodes.times do
  id, left_id, right_id = gets.split.map &:to_i
  left  = left_id  == -1 ? Node::NIL : tree[left_id]
  right = right_id == -1 ? Node::NIL : tree[right_id]
  Node.bind tree[id], left, right
end

tree.find { |node| node.root? }.update_tree

tree.each do |n|
  puts "node #{n.id}: parent = #{n.parent.id}, " <<
       "sibling = #{n.sibling.id}, " <<
       "degree = #{n.degree}, depth = #{n.depth}, " <<
       "height = #{n.height}, #{n.type}"
end