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: WA: Presentation Error)


class Tree
	attr_accessor :parent, :root, :dep, :key, :left,:right, :sib
	def initialize(key)
		@parent = -1
		@left = -1
		@right = -1
		@dep = -1
		@key = key
		@sib = -1
	end

	def depth(ans,tree)
		if @left!=-1
			@left.depth(ans+1,tree)
		end
		
		if @right != -1
			@right.depth(ans+1,tree)
		end
		@dep = ans
	end

	def height(h)
		n1 = h
		n2 = h

		if @left!=-1
			n1 = left.height(h+1) 
		end

		if @right!=-1
			n2 = right.height(h+1)
		end

		return n1 > n2 ? n1:n2
	end

	def showparent
		if @parent==-1
			return -1
		else
			return @parent.key
		end
	end

	def showsib
		if @sib == -1
			return -1
		else
			return @sib.key
		end
	end

	def degree
		ans = 0
		if @left != -1
			ans+=1
		end
		if @right != -1
			ans+=1
		end
		ans
	end

	def preorder(list)
		list << @key
		if @left != -1
			list = @left.preorder(list)
		end
		if @right != -1
			list = @right.preorder(list)
		end
		
		list
	end	

	def inorder(list)
		if @left != -1
			list = @left.inorder(list)
		end
		list << @key
		if @right != -1
			list = @right.inorder(list)
		end

		list
	end

	def postorder(list)
		if @left != -1
			list = @left.postorder(list)
		end

		if @right != -1
			list = @right.postorder(list)
		end
		list << @key
		list
	end
end




n = gets.to_i

tree = Hash.new
check = Array.new(n){true}
n.times do |i|
	tree[i] = Tree.new(i)
end

n.times do
	data = gets.split.map(&:to_i)
	num = data[0]
	tree[num].left = tree[data[1]]||-1
	tree[num].right = tree[data[2]]||-1
	if tree[data[1]]
		tree[data[1]].parent = tree[num]
		tree[data[1]].sib = tree[data[2]]||-1
		check[data[1]] = false
	end
	if tree[data[2]]
		tree[data[2]].parent = tree[num]
		tree[data[2]].sib = tree[data[1]]||-1
		check[data[2]] = false
	end
	
	
end
r = check.index(true)
obj = tree[r]
list = []
puts "Preorder"
puts obj.preorder(list).join(" ")
list = []

puts "Inorder"
puts obj.inorder(list).join(" ")
list = []

puts "Postorder"
puts obj.postorder(list).join(" ")