Print nodes in top view of Binary Tree | Set 2

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 should be printed from left to right.

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

Input:
       1
    /     \
   2       3
  /  \    / \
 4    5  6   7
Output: Top view: 4 2 1 3 7

Input:
        1
      /   \
    2       3
      \   
        4  
          \
            5
             \
               6
Output: Top view: 2 1 3 6

The idea is to do something similar to Vertical Order Traversal. Like Vertical Order Traversal, we need to group nodes of same horizontal distance together. We do a level order traversal so that the topmost node at a horizontal node is visited before any other node of same horizontal distance below it. A Map is used to map the horizontal distance of the node with the node’s Data and vertical distance of the node.

Below is the implementation of the above approach:

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// C++ Program to print Top View of Binary Tree
// using hashmap and recursion
#include <bits/stdc++.h>
using namespace std;
  
// Node structure
struct Node {
    // Data of the node
    int data;
  
    // Horizontal Distance of the node
    int hd;
  
    // Reference to left node
    struct Node* left;
  
    // Reference to right node
    struct Node* right;
};
  
// Initialising node
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->hd = INT_MAX;
    node->left = NULL;
    node->right = NULL;
    return node;
}
  
void printTopViewUtil(Node* root, int height,
    int hd, map<int, pair<int, int> >& m)
{
    // Base Case
    if (root == NULL)
        return;
  
    // If the node for particular horizontal distance
    // is not present in the map, add it.
    // For top view, we consider the first element 
    // at horizontal distance in level order traversal
    if (m.find(hd) == m.end()) {
        m[hd] = make_pair(root->data, height);
    }
    else{
        pair<int, int> p = (m.find(hd))->second;
                  
        if (p.second > height) {
            m.erase(hd);
            m[hd] = make_pair(root->data, height);
        }
    }
  
    // Recur for left and right subtree
    printTopViewUtil(root->left, height + 1, hd - 1, m);
    printTopViewUtil(root->right, height + 1, hd + 1, m);
}
  
void printTopView(Node* root)
{
    // Map to store horizontal distance, 
    // height and node's data
    map<int, pair<int, int> > m;
    printTopViewUtil(root, 0, 0, m);
  
    // Print the node's value stored by printTopViewUtil()
    for (map<int, pair<int, int> >::iterator it = m.begin(); 
                                        it != m.end(); it++) {
        pair<int, int> p = it->second;
        cout << p.first << " ";
    }
}
  
int main()
{
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->right = newNode(4);
    root->left->right->right = newNode(5);
    root->left->right->right->right = newNode(6);
  
    cout << "Top View : ";
    printTopView(root);
  
    return 0;
}

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Output:


Top View : 2 1 3 6






          
          
          
          

            

    

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