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Find top three elements of binary tree in scala

Finding top 3 maximum node in binary tree

Scala program for Find top three elements of binary tree. Here mentioned other language solution.

/* 
  Scala program for
  Find top three elements in binary tree
*/
// Node of Binary Tree
class TreeNode(var data: Int,
	var left: TreeNode,
		var right: TreeNode)
{
	def this(data: Int)
	{
		// Set node value
		this(data, null, null);
	}
}
class BinaryTree(var root: TreeNode,
	var first: TreeNode,
		var second: TreeNode,
			var third: TreeNode)
{
	def this()
	{
		this(null, null, null, null);
	}
	// Recursive function
	// Display preorder view of binary tree
	def preOrder(node: TreeNode): Unit = {
		if (node != null)
		{
			//Print node value
			print("  " + node.data);
			preOrder(node.left);
			preOrder(node.right);
		}
	}
	// Find top three largest nodes in binary tree
	def getTopThree(node: TreeNode): Unit = {
		if (node != null)
		{
			if (this.first == null)
			{
				// First node of tree
				this.first = node;
			}
			else if (this.first.data < node.data)
			{
				// When get new first largest node
				// Interchange the node values
				this.third = this.second;
				this.second = this.first;
				this.first = node;
			}
			else if (this.second == null)
			{
				if (this.first.data != node.data)
				{
					// Get second largest node
					this.second = node;
				}
			}
			else
			{
				if (this.second.data < node.data && 
                    this.first.data > node.data)
				{
					// When we get new second largest
					// Replace the third node with the second node
					this.third = second;
					// This is new second largest element
					this.second = node;
				}
				else if (this.third == null)
				{
					if (this.second.data > node.data && 
                        this.first.data > node.data)
					{
						// Get the third largest node
						this.third = node;
					}
				}
				else if (this.third.data < node.data && 
                         this.first.data > node.data && 
                         this.second.data > node.data)
				{
					this.third = node;
				}
			}
			// Recursively, execute left and right subtree
			getTopThree(node.left);
			getTopThree(node.right);
		}
	}
	// This are handle the request to finding
	// top three nodes in binary tree.
	def printTopThree(): Unit = {
		if (this.root == null)
		{
			return;
		}
		// Display tree elements
		print("\n Preorder : ");
		preOrder(this.root);
		// Set the initial all three nodes
		this.first = null;
		this.second = null;
		this.third = null;
		// Find three largest element
		getTopThree(this.root);
		// Display calculated result
		println("\n First : " + first.data);
		if (second != null && 
            second != first && 
            second.data < first.data)
		{
			println(" Second : " + second.data);
		}
		else
		{
			println(" Second : None ");
		}
		if (third != null && third != second && 
            third != first && third.data < second.data)
		{
			println(" Third : " + third.data);
		}
		else
		{
			println(" Third : None");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		// Create two trees
		var x: BinaryTree = new BinaryTree();
		var y: BinaryTree = new BinaryTree();
		/*
		    Construct Binary Tree
		    -----------------------
		           10
		          /  \
		         /    \
		        6      8
		       / \    / \
		      12  3  7   5
		     /        \   \
		    9         -6  13
		*/
		// Add nodes
		x.root = new TreeNode(10);
		x.root.left = new TreeNode(6);
		x.root.left.left = new TreeNode(12);
		x.root.right = new TreeNode(8);
		x.root.right.right = new TreeNode(5);
		x.root.right.left = new TreeNode(7);
		x.root.right.left.right = new TreeNode(-6);
		x.root.left.right = new TreeNode(3);
		x.root.left.left.left = new TreeNode(9);
		x.root.right.right.right = new TreeNode(13);
		/*
		    Construct Tree
		    --------------
		         1
		        / \ 
		       /   \
		      1     2
		     /     / \
		    1     2   2
		*/
		// Add second tree node
		y.root = new TreeNode(1);
		y.root.left = new TreeNode(1);
		y.root.right = new TreeNode(2);
		y.root.right.right = new TreeNode(2);
		y.root.right.left = new TreeNode(2);
		y.root.left.left = new TreeNode(1);
		// Test One
		x.printTopThree();
		// Test Two
		y.printTopThree();
	}
}

Output

 Preorder :   10  6  12  9  3  8  7  -6  5  13
 First : 13
 Second : 12
 Third : 10

 Preorder :   1  1  1  2  2  2
 First : 2
 Second : 1
 Third : None




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