Check for foldable binary tree in ruby

Ruby program for Check for foldable binary tree. Here problem description and explanation.

#  Ruby program for
#  Check if binary tree is foldable binary tree
#  Using recursion

#  Binary Tree Node
class TreeNode 
	# Define the accessor and reader of class TreeNode
	attr_reader :data, :left, :right
	attr_accessor :data, :left, :right
	def initialize(data) 
		#  Set node value
		self.data = data
		self.left = nil
		self.right = nil
	end
end

class BinaryTree 
	# Define the accessor and reader of class BinaryTree
	attr_reader :root
	attr_accessor :root
	def initialize() 
		self.root = nil
	end
	def isFoldable(leftRoot, rightRoot) 
		if (leftRoot == nil && rightRoot == nil) 
			#  When both node empty
			return true
		elsif(leftRoot != nil && rightRoot != nil &&
              self.isFoldable(leftRoot.left, rightRoot.right) && 
              self.isFoldable(leftRoot.right, rightRoot.left)) 
			#  When
			#  Valid the properties of foldable tree
			return true
		end
		#  Fail case
		return false
	end

	#  Display tree element inorder form
	def inorder(node) 
		if (node != nil) 
			#  Visit to left subtree
			self.inorder(node.left)
			#  Print node value
			print("  ", node.data)
			#  Visit to right subtree
			self.inorder(node.right)
		end
	end
end

def main() 
	#  Create new tree
	tree = BinaryTree.new()
	#   Binary Tree
	#  -----------------------
	#       5
	#     /   \
	#    7     4
	#   /       \
	#  1         2
	#   \       /
	#    2     5
	#  Binary tree nodes
	tree.root = TreeNode.new(5)
	tree.root.left = TreeNode.new(7)
	tree.root.right = TreeNode.new(4)
	tree.root.left.left = TreeNode.new(1)
	tree.root.right.right = TreeNode.new(2)
	tree.root.right.right.left = TreeNode.new(5)
	tree.root.left.left.right = TreeNode.new(2)
	#  Display Tree elements
	tree.inorder(tree.root)
	result = tree.isFoldable(tree.root.left, 
                             tree.root.right)
	if (result == true) 
		print("\n Foldable Tree\n")
	else
		print("\n Not Foldable Tree\n")
	end

	#         5
	#       /   \
	#      7     4
	#     /       \
	#    1         2
	#     \       /
	#      2     5
	#       \
	#        3  
	#  Case 2
	tree.root.left.left.right.right = TreeNode.new(3)
	tree.inorder(tree.root)
	result = tree.isFoldable(tree.root.left, 
                             tree.root.right)
	if (result == true) 
		print("\n Foldable Tree\n")
	else
		print("\n Not Foldable Tree\n")
	end

end

main()

Output

  1  2  7  5  4  5  2
 Foldable Tree
  1  2  3  7  5  4  5  2
 Not Foldable Tree