spiralTravese function in the given programme takes up two stacks s1 and s2 and print even level from right to left and odd level from left to right,level of a binary tree starts with level 0. Time complexity is O(n).

```
#include<bits/stdc++.h>
using namespace std;
struct node{
struct node *left;
struct node *right;
int data;
};
typedef struct node node;
node *newNode(int data){
node *temp=(node*)malloc(sizeof(node));
temp->left=temp->right=NULL;
temp->data=data;
return temp;
}
void spiralTraverse(node *root){
stack <node *> s1,s2;
s1.push(root);
while(!s1.empty() || !s2.empty()){
while(!s1.empty()){
node *tmp=s1.top();
s1.pop();
printf("%d ",tmp->data);
if(tmp->right)
s2.push(tmp->right);
if(tmp->left)
s2.push(tmp->left);
}
while(!s2.empty()){
node *tmp=s2.top();
printf("%d ",tmp->data);
s2.pop();
if(tmp->left)
s1.push(tmp->left);
if(tmp->right)
s1.push(tmp->right);
}
}
}
int main(){
node *root=NULL;
root=newNode(1);
root->left=newNode(2);
root->right=newNode(3);
root->left->left=newNode(4);
root->left->right=newNode(5);
root->right->left=newNode(6);
root->right->right=newNode(7);
spiralTraverse(root);
return 0;
}
```