Top-View.cpp

/*
 
 QUES :  Top view of a binary tree is the set of nodes visible when the tree is viewed from the top. Given a binary tree, print the top view of it. The output nodes can be    printed  in any order.

        A node x is there in output if x is the topmost node at its horizontal distance. Horizontal distance of left child of a node x is equal to horizontal distance of x minus 1, and that of right child is horizontal distance of x plus 1.
*/

/*
 Approach:
      
      1.We will create 2 maps map1 and map2,first we will store the position  corresponding to an node and the other will store the node corresponding to a
        particular position.
      2.And we will use level order traversal that if it's left/right root exist then it will push in the queue and stores it's pos in the
        map1. for left node position will become pos-1 and for right node position will become pos+1.
      3.Doing this work till queue does not become empty;
      4.While poping the element ,find the pos corresponding to it .If the key as the pos is already present in map2 don't do anything otherwise
        place it in map2;
      5.After this whole thing we will have map2 with the unique postions and the node's value corresponding to it and in the ascending order
        as well
      6.So just iterate through the map2 and print the second value...
 
 */


#include<iostream>
using namespace std;
#include <map>

//creating a Binary Tree Node class 
template <typename T>  //makes sure that we can make a node of any data type we want 
class BinaryTreeNode {
    public : 
    T data;                      //data corresponding to the paticular node 
    BinaryTreeNode<T> *left;     //stores the address of the left side of the BT
    BinaryTreeNode<T> *right;    //stores the address of the right side of the BT

    BinaryTreeNode(T data) {
        this -> data = data;
        left = NULL;
        right = NULL;
    }
};

//Take the Input Of the Binary Tree in the level Order Traversal 
// Make sure if there is left or right node of any node then give it's data as -1;

BinaryTreeNode< int >* takeInput() {
    
    int rootData;
    cin >> rootData;
    if(rootData == -1) {
        return NULL;
    }
    //the root node
    
    BinaryTreeNode< int > *root = new BinaryTreeNode< int >( rootData );
    queue<BinaryTreeNode< int >*> q;
    q.push( root );
    
    while(!q.empty())
    {
        
        BinaryTreeNode< int > *currentNode = q.front();
	    q.pop();
        int leftChild, rightChild ;
        
        //giving the data for the leftChild .If the data provided is -1  don't make any node of it .
        
        cin >> leftChild;
        
        if( leftChild != -1 ) {
            BinaryTreeNode< int >* leftNode = new BinaryTreeNode< int >( leftChild );
            currentNode -> left = leftNode;
            q.push( leftNode );
        }
        
        //giving the data for the leftChild .If the data provided is -1  don't make any node of it .
        
        cin >> rightChild;
        
        if( rightChild != -1 ) {
            BinaryTreeNode< int >* rightNode = new BinaryTreeNode< int >( rightChild ) ;
            currentNode -> right = rightNode;
            q.push( rightNode );
        }
    }
    
    //returning the root of the Binary Tree
    
    return root;
}

void TopViewofBT( BinaryTreeNode< int >* root )
{
	//Base Case
	if(root == NULL)
	{
		return ;
	}
    //stores the pos corresponding to a node .Here Key:Node and value is position
    
	map < Node*, int > m;
    
    //stores the node's value corresponding to a pos .Here Key:position and value is Node's value
    
    map < int , int > mp;
    
    queue< Node* >q;
    
    //initializing the root's position as 0;
    m[root] = 0;
    
    q.push( root );
    
    while( !q.empty() )
    {
        Node* temp = q.front();
        q.pop();
        
        //getting the pos of the particular node
        
        int t = m[temp];
        
        if(mp.find(t) == mp.end())
        {
            //storing the node's value corresponding to the unique pos
            mp[t] = temp->data;
        }
        
        //checking if node's left exist
        
        if( temp->left )
        {
            //pos_left = pos-1
            m[temp->left] = t - 1;
            q.push(temp->left);
        }
        
        //checking if node's right exist
        
        if(temp->right)
        {
            //pos_right = pos+1;
            q.push( temp->right );
            m[temp->right] = t + 1;
        }
    }
     //iterating through the map
    
    for( map<int,int>::iterator i = mp.begin() ; i != mp.end() ; i++)
    {
        //printing the node's value
        cout<< i->second <<" ";
    }
    
    cout<< endl;
}

int main()
{
    
    BinaryTreeNode<int>* root = takeInput();
    
    TopViewofBT(root);
    
}

/*
 
 TIME COMPLEXITY : O(nlogn) , n is the number of nodes in a Binary Tree
 
 SPACE COMPLEXITY : O(n) , n is the number of nodes in a Binary Tree
 
 
 EXAMPLES:
 
 1.
 
          1
        2   3
      4  5 6  7

 
    INPUT  : 1 2 3 4 5 6 7 -1
 
    OUTPUT : 4 2 1 3 7
 
 
 2.
 
 
     1
   /   \
 2       3
   \
     4
       \
         5
          \
            6
 
 
   INPUT  : 1 2 3 -1 4 -1 -1 -1 5 -1 6
   
   OUTPUT : 2 1 3 6
 
 
 */