Binary Tree -Zig Zag Order

[SharedMocha]

//https://www.geeksforgeeks.org/level-order-traversal-in-spiral-form/
//https://leetcode.com/explore/interview/card/top-interview-questions-medium/108/trees-and-graphs/787/

// Java implementation of an O(n) approach of level order
// traversal in spiral form

import java.util.*;

// A Binary Tree node
class Node
{
int data;
Node left, right;

public Node(int item) 
{ 
	data = item; 
	left = right = null; 
} 

}

public class BinaryTree
{

static Node root; 

void printSpiral(Node node) 
{ 
 if(node == null){
 return;
 }
 Stack<Node> one = new Stack<Node>();
 Stack<Node> two = new Stack<Node>();
 one.push(node);
 while(!one.empty() || !two.empty()){
 
 while(!one.empty()){
 
 Node oneTemp = one.peek();
 one.pop();
 if(oneTemp != null){
 System.out.println(oneTemp.data);
 if(oneTemp.right != null){
 two.push(oneTemp.right);
 }
 if(oneTemp.left != null){
 two.push(oneTemp.left);
 
 } 
 }
 
 
 }
 
 while(!two.empty()){
 Node twoTemp = two.peek();
 two.pop();
 System.out.println(twoTemp.data);
 if(twoTemp != null){
 if(twoTemp.left != null){
 one.push(twoTemp.left);
 }
 if(twoTemp.left != null){
 one.push(twoTemp.right);
 
 } 
 } 
 }
 
 }
} 

public static void main(String[] args) 
{ 
	BinaryTree tree = new BinaryTree(); 
	tree.root = new Node(1); 
	tree.root.left = new Node(2); 
	tree.root.right = new Node(3); 
	tree.root.left.left = new Node(7); 
	tree.root.left.right = new Node(6); 
	tree.root.right.left = new Node(5); 
	tree.root.right.right = new Node(4); 
	System.out.println("Spiral Order traversal of Binary Tree is "); 
	tree.printSpiral(root); 
} 

}

// This code has been contributed by Mayank Jaiswal(mayank_24)