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)

class Node

  attr_accessor:id, :parent, :left, :right, :depth, :height, :sibling

  def initialize(id)
    @id = id
    @parent = nil
    @left = nil
    @right = nil
    @depth = 0
    @height = 0
    @sibling = nil
  end

  def depth
    if @parent.nil?
      0
    else
      @parent.depth + 1
    end
  end

  def type
    if @parent == nil
      "root"
    elsif @child_left.nil?
      "leaf"
    else
      "internal node"
    end
  end

  def to_s
    parent_id =
      if @parent.nil?
        -1
      else
        @parent.id
      end

    sibling =
      if @sibling.nil?
        -1
      else
        @sibling.id
      end

    degree = 0
    if @left != nil
      degree = degree + 1
    end
    if @right != nil
      degree = degree + 1
    end

    type =
      if @parent.nil?
        "root"
      elsif @left.nil? && @right.nil?
        "leaf"
      else
        "internal node"
      end

    "node #{@id}: parent = #{parent_id}, sibling = #{sibling}, degree = #{degree}, depth = #{depth}, height = #{height}, #{type}"
  end

  def children_to_s
    ss = []
    if @child_left.nil?
      ss
    else
      a = @child_left
      while a != nil
        ss.push(a.id.to_s)
        a = a.right_sibling
      end
      ss
    end
  end
end

n = STDIN.gets.to_i

tree = []
for i in 0..(n - 1)
  as = STDIN.gets.split(' ')
  id = as[0].to_i
  left_id = as[1].to_i
  right_id = as[2].to_i

  node = tree[id]
  if node.nil?
    node = Node.new(id)
    tree[id] = node
  end

  if left_id != -1
    left = tree[left_id]
    if left.nil?
      left = Node.new(left_id)
      tree[left_id] = left
    end
    left.parent = node
    node.left = left
  end

  if right_id != -1
    right = tree[right_id]
    if right.nil?
      right = Node.new(right_id)
      tree[right_id] = right
    end
    right.parent = node
    node.right = right
  end

  if left_id != -1 && right_id != -1
    node.left.sibling = node.right
    node.right.sibling = node.left
    a = node
    while a != nil
      a.height = a.height + 1
      a = a.parent
    end
  end
end

for i in 0..(n-1)
  puts tree[i].to_s
end