spiral traversal of binary tree

This can be done using two stacks in iterative manner.

The below diagram depicts the spiral traversal of BST
SpriralBST

Steps to solve
push level_1 nodes (i.e, only root node) into stack_1;
next by popping elements of level_1, push children of level_l in to stack_2 (i.e, level_2);
next by popping elements of level_2, push children of level_2 in to stack_2 (i.e, level_3);
… so on

Code:

	/**
	 * Algorithm to print binary tree in spiral manner
	 * 
	 * @param node root node of the binary tree
	 */
	public void printTreeInSpiralForm(BinaryNode node) {
		Stack<BinaryNode> stack1 = new Stack<BinaryNode>();
		Stack<BinaryNode> stack2 = new Stack<BinaryNode>();
		stack1.push(node);

		while (!stack1.empty() || !stack2.empty()) {
			//
			while (!stack1.empty()) {
				node = stack1.pop();
				if(node != null) {
					System.out.print(node.element+", ");
					if(node.left != null)
						stack2.push(node.left);
					if(node.right != null)
						stack2.push(node.right);
				}
			}

			while (!stack2.empty() && node != null) {
				node = stack2.pop();
				if(node != null) {
					System.out.print(node.element+", ");
					if(node.left != null)
						stack1.push(node.right);
					if(node.right != null)
						stack1.push(node.left);
				}
			}
		}
	}

Output:
20, 25, 15, 12, 18, 22, 28, 32, 26, 24, 21, 19, 17, 14, 6

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