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: Accepted)

n = gets.to_i

class Node
  attr_accessor :id, :parent, :left, :right, :sibling, :degree, :depth, :height, :node_type
  def initialize(i)
    @id = i
    @parent = -1
    @left = -1
    @right = -1
    @depth = 0
    @height = 0
    @sibling = -1
    @degree = -1
  end
end

nodes = {}
# 大きさが分かるので初期化しておく
n.times do |i|
  nodes[i] = Node.new(i)
end

# 高さを決める
def height_rec(par, nodes)
  l, r = 0, 0
  if nodes[par].left >= 0
    tmp_left = nodes[par].left
    l = height_rec(tmp_left, nodes) + 1
  end
  if nodes[par].right >= 0
    tmp_right = nodes[par].right
    r = height_rec(tmp_right, nodes) + 1
  end
  nodes[par].height = [r, l].max
end

# 深さを再帰的に更新
def depth_rec(par, nodes)
  left_child = nodes[par].left
  if left_child >= 0
    nodes[left_child].depth = nodes[par].depth + 1
    depth_rec(left_child, nodes)
  end
  right_child = nodes[par].right
  if right_child >= 0
    nodes[right_child].depth = nodes[par].depth + 1
    depth_rec(right_child, nodes)
  end
end

n.times do |i|
  data = gets.chomp.split.map(&:to_i)
  my_id = data[0]
  child_l = data[1]
  child_r = data[2]
  # 子データをセット
  nodes[my_id].left = child_l
  nodes[my_id].right = child_r
  # 子に親と兄弟データをセット
  if child_l >= 0
    nodes[child_l].parent = my_id
    if child_r >= 0
      nodes[child_l].sibling = child_r
    end
  end
  if child_r >= 0
    nodes[child_r].parent = my_id
    if child_l >= 0
      nodes[child_r].sibling = child_l
    end
  end
  # 子の数をセット
  nodes[my_id].degree = [child_l, child_r].select{|c|c!=-1}.size
  # 深さ更新
  depth_rec(my_id, nodes)
end

# 高さ計算とタイプ確定
# root、internal node、leaf 
nodes.each do |k, node|
  height_rec(k, nodes)
  if node.parent == -1
    node.node_type = "root"
  elsif node.degree.zero? 
    node.node_type = "leaf"
  else
    node.node_type = "internal node"
  end
end

# 出力
# node id: parent = p , sibling = s , degree = deg, depth = dep, height = h, type
nodes.each do |k, node|
  puts "node #{node.id}: parent = #{node.parent}, sibling = #{node.sibling}, degree = #{node.degree}, depth = #{node.depth}, height = #{node.height}, #{node.node_type}"
end