In this article I will show you how to traverse a tree in zigzag manner
// zigzag trav for tree
>
// 12
// / \
<
// 9 15
// / \
>
// 4 10
// output : 12 15 9 4 10
#include<iostream>
#include<stack>
using namespace std;
class Node
{
public:
int data;
Node *right,*left;
Node(int data)
{
this>data=data;
right=left=NULL;
}
};
void zigzagtrav(Node *root)
{
stack<Node*> curr;
stack<Node*> next;
bool leftToright=true;
curr.push(root);
while(!curr.empty())
{
Node *tmp=curr.top();
curr.pop();
if(tmp)
{
cout<<tmp>data<<" ";
if(leftToright)
{
next.push(tmp>left);
next.push(tmp>right);
}
else
{
next.push(tmp>right);
next.push(tmp>left);
}
}
if(curr.empty())
{
leftToright=!leftToright;
swap(curr,next);
}
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("../input.txt", "r", stdin);
freopen("../output.txt", "w", stdout);
#endif
Node *root=new Node(12);
root>right=new Node(15);
root>left=new Node(9);
root>left>left=new Node(4);
root>left>right=new Node(10);
zigzagtrav(root);
return 0;
}
here we are using two stacks

curr
which is to store our current node 
next
which is to store the next node  here the
leftToright
is to check that are we are printing the nodes lefttoright or righttoleft if the variable is true then we are printing the nodes left to right else print righttoleft.  Here if
leftToright
is true we are pushing the current left and then right node to next stack else push current right and then left node to next stack.  and checking if the curr node is empty swap the stack with next stack and toggle the
leftToright
variable.
Do it until both stack is empty.
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