Problem:
Tree Walk

Binary trees are defined recursively. A binary tree T is a structure defined on a finite set of nodes that either

contains no nodes, or

is composed of three disjoint sets of nodes:

- a root node.

- a binary tree called its left subtree.

- a binary tree called its right subtree.

Your task is to write a program which perform tree walks (systematically traverse all nodes in a tree) based on the following algorithms:

Print the root, the left subtree and right subtree (preorder).

Print the left subtree, the root and right subtree (inorder).

Print the left subtree, right subtree and the root (postorder).

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 linen, 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

In the 1st line, print "Preorder", and in the 2nd line print a list of node IDs obtained by the preorder tree walk.

In the 3rd line, print "Inorder", and in the 4th line print a list of node IDs obtained by the inorder tree walk.

In the 5th line, print "Postorder", and in the 6th line print a list of node IDs obtained by the postorder tree walk.

Print a space character before each node ID.

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

Preorder
0 1 2 3 4 5 6 7 8
Inorder
2 1 3 0 6 5 7 4 8
Postorder
2 3 1 6 7 5 8 4 0

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)

class MyBinaryNode
  attr_accessor :parent, :left, :right

  def initialize
    @parent = -1
    @left = -1
    @right = -1
  end
end

def pre_order(nodes, id)
  if id == -1
    return 
  end
  print ' '
  print id
  pre_order(nodes, nodes[id].left)
  pre_order(nodes, nodes[id].right)
end

def in_order(nodes, id)
  if id == -1
    return 
  end
  in_order(nodes, nodes[id].left)
  print ' '
  print id
  in_order(nodes, nodes[id].right)
end

def post_order(nodes, id)
  if id == -1
    return 
  end
  post_order(nodes, nodes[id].left)
  post_order(nodes, nodes[id].right)
  print ' '
  print id
end

n = gets.to_i
nodes = []
n.times do
  nodes << MyBinaryNode.new
end

n.times do
  inputs = gets.split.map(&:to_i)
  node_id = inputs.shift
  node = nodes[node_id]
  left = inputs.shift
  if left != -1
    node.left = left
    left_node = nodes[left]
    left_node.parent = node_id
  end
  right = inputs.shift
  if right != -1
    node.right = right
    right_node = nodes[right]
    right_node.parent = node_id
  end
end

root_id = nil
nodes.each_with_index do |node, i|
  if node.parent == -1
    root_id = i
  end
end

puts "Preorder"
pre_order(nodes, root_id)
puts ""

puts "Inorder"
in_order(nodes, root_id)
puts ""

puts "Postorder"
post_order(nodes, root_id)
puts ""