# Check for foldable binary tree in swift

Swift program for Check for foldable binary tree. Here problem description and other solutions.

``````import Foundation
/*
Swift 4 program for
Check if binary tree is foldable binary tree
Using recursion
*/
// Binary Tree Node
class TreeNode
{
var data: Int;
var left: TreeNode? ;
var right: TreeNode? ;
init(_ data: Int)
{
// Set node value
self.data = data;
self.left = nil;
self.right = nil;
}
}
class BinaryTree
{
var root: TreeNode? ;
init()
{
self.root = nil;
}
func isFoldable(_ leftRoot: TreeNode? ,
_ rightRoot : TreeNode? ) -> Bool
{
if (leftRoot == nil && rightRoot == nil)
{
// When both node empty
return true;
}
else if (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;
}
// Fail case
return false;
}
// Display tree element inorder form
func inorder(_ node: TreeNode? )
{
if (node  != nil)
{
// Visit to left subtree
self.inorder(node!.left);
// Print node value
print("",node!.data, terminator: " ");
// Visit to right subtree
self.inorder(node!.right);
}
}
static func main()
{
// Create new tree
let tree: BinaryTree = BinaryTree();
/*
Binary Tree
-----------------------
5
/   \
7     4
/       \
1         2
\       /
2     5
*/
// Binary tree nodes
tree.root = TreeNode(5);
tree.root!.left = TreeNode(7);
tree.root!.right = TreeNode(4);
tree.root!.left!.left = TreeNode(1);
tree.root!.right!.right = TreeNode(2);
tree.root!.right!.right!.left = TreeNode(5);
tree.root!.left!.left!.right = TreeNode(2);
// Display Tree elements
tree.inorder(tree.root);
var result: Bool = tree.isFoldable(
tree.root!.left, tree.root!.right);
if (result == true)
{
print("\n Foldable Tree");
}
else
{
print("\n Not Foldable Tree");
}
/*
5
/   \
7     4
/       \
1         2
\       /
2     5
\
3
*/
// Case 2
tree.root!.left!.left!.right!.right = TreeNode(3);
tree.inorder(tree.root);
result = tree.isFoldable(
tree.root!.left, tree.root!.right);
if (result == true)
{
print("\n Foldable Tree");
}
else
{
print("\n Not Foldable Tree");
}
}
}
BinaryTree.main();``````

Output

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

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