spiral traversal of binary tree

This can be done using two stacks in iterative manner.

The below diagram depicts the spiral traversal of BST

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


	 * 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>();

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

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

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


5 Responses to spiral traversal of binary tree

  1. sreeram says:

    A single queue could also be used .

    Initially Queue would be populated with root. A loop would iterate on the queue ,print the node ,pick a node add its right child and left child to the queue and so on till no elements are there in the queue.

  2. @sreeram: doesn’t work.

  3. it should work, send me the complete code

  4. ankit says:

    If node.left!=null. Why pushing node.right in stack1 ? And why stack2.empty== false instead of simply !stack2.empty()

  5. ankit says:

    Also we are never pushing null in stacks…so why checking for nulls ?

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