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)

#binary_tree
class Node
  attr_accessor :myname,:parent,:left,:right,:depth,:height

  def initialize(n)
    @myname = n
    @parent = nil
    @left   = nil
    @right  = nil
    @depth  = 0
  end

  def add_left(child)
      self.left    = child
      child.parent = self
  end

  def add_right(child)
      self.right    = child
      child.parent = self
  end

  def set_depth(d)
    self.depth = d
    if(self.left != nil)
      self.left.set_depth(d+1)
    end
    if(self.right != nil)
      self.right.set_depth(d+1)
    end
  end

  def set_height
    if(self.left != nil)
      lh=self.left.set_height
    else
      lh=0
    end
    if(self.right != nil)
      rh=self.right.set_height
    else
      rh=0
    end
    if lh > rh 
      self.height = lh + 1
    else
      self.height = rh + 1
    end
  end

  def sibling
    p = self.parent
    if p == nil
      nil
    elsif self == p.left
      p.right
    else
      p.left
    end
  end

  def degree
    deg=0
    if self.left != nil
      deg += 1
    end
    if self.right != nil
      deg += 1
    end
    deg
  end

  def shubetu

    if self.parent == nil
      "root"
    elsif self.height == 1
      "leaf"
    else
      "internal node"
    end

  end

  def describe
    printf("node %d:",self.myname)
    if self.parent == nil 
      printf(" parent = -1,")
    else
      printf(" parent = %d,",self.parent.myname)
    end
    if self.sibling == nil 
      printf(" sibling = -1,")
    else
      printf(" sibling = %d,",self.sibling.myname)
    end
    printf(" degree = %d,",self.degree)
    printf(" depth = %d,",self.depth)
    printf(" height = %d,",self.height-1)
    printf(" %s\n",self.shubetu)
  end
end


$n = STDIN.gets.split[0].to_i
$tree=Array.new($n+1){|i| Node.new(i)}

0.upto($n-1){|i|
  s = STDIN.gets.split
  id= s[0].to_i
  l = s[1].to_i
  r = s[2].to_i

#  printf("%d %d %d:\n",id,l,r)
  if(l != -1)
    $tree[id].add_left($tree[l])
  end
  if(r != -1)
    $tree[id].add_right($tree[r])
  end
}

def findroot
  0.upto($n-1){|i|
    if $tree[i].parent == nil 
      return i
    end
  }
end

rt = findroot
printf("root=%d\n",rt)

$tree[rt].set_depth(0)
$tree[rt].set_height

0.upto($n-1){|i|
  $tree[i].describe 
}