# 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

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

### 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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