Check if all leaf nodes are at same level in kotlin

Kotlin program for Check if all leaf nodes are at same level. Here problem description and other solutions.

/* 
  Kotlin program for
  Check if all leaves are at same level 
*/

// Binary Tree Node
class TreeNode
{
	var data: Int;
	var left: TreeNode ? ;
	var right: TreeNode ? ;
	constructor(data: Int)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinaryTree
{
	var root: TreeNode ? ;
	var depth: Int;
	constructor()
	{
		this.root = null;
		this.depth = -1;
	}
	fun checkLeafHeight(node: TreeNode ? , level : Int): Unit
	{
		if (node != null)
		{
			// Test node is leaf or not
			if (node.left == null && node.right == null)
			{
				if (this.depth == -1)
				{
					// This is first leaf node
					// Get node level
					this.depth = level;
				}
				else if (this.depth != level)
				{
					// When two leaf nodes is different level
					this.depth = -2;
					return;
				}
			}
			if (this.depth != -2)
			{
				// Recursively visit left and right subtree
				this.checkLeafHeight(node.left, level + 1);
				this.checkLeafHeight(node.right, level + 1);
			}
		}
	}
	fun isSameLevelLeaf(): Unit
	{
		this.depth = -1;
		// Test
		this.checkLeafHeight(this.root, 1);
		// Display result
		if (this.depth < 0)
		{
			// Two possible case
			// 1) When tree empty
			// 2) When leaf node are not same level
			println("No");
		}
		else
		{
			// When leaf node is same level
			println("Yes");
		}
	}
}
fun main(args: Array < String > ): Unit
{
	// Create new tree
	val tree: BinaryTree = BinaryTree();
	/*   
	    Make A Binary Tree
	    --------------------
	      1
	     /  \
	    2    3
	   /    /  \
	  4    5    6
	       
	*/
	//insertion of binary Tree nodes
	tree.root = TreeNode(1);
	tree.root?.left = TreeNode(2);
	tree.root?.right = TreeNode(3);
	tree.root?.right?.right = TreeNode(6);
	tree.root?.right?.left = TreeNode(5);
	tree.root?.left?.left = TreeNode(4);
	tree.isSameLevelLeaf();
	// Add new node
	tree.root?.left?.left?.left = TreeNode(7);
	/*
	   When add new node
	    -----------------------
	          1
	         /  \
	        2    3
	       /    /  \
	      4    5    6
	     /
	    7  
	*/
	tree.isSameLevelLeaf();
}

Output

Yes
No