Print nodes in the Top View of Binary Tree | Set 3

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. Expected time complexity is O(n)

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. 

Example :

       1
    /     \
   2       3
  /  \    / \
 4    5  6   7
Top view of the above binary tree is
4 2 1 3 7

        1
      /   \
    2       3
      \   
        4  
          \
            5
             \
               6
Top view of the above binary tree is
2 1 3 6

Approach:

• The idea here is to observe that, if we try to see a tree from its top, then only the nodes which are at top in vertical order will be seen.
• Start BFS from root. Maintain a queue of pairs comprising of node(Node *) type and horizontal distance of node from root. Also, maintain a map which should store the node at a particular horizontal distance.
• While processing a node, just check if any node is there in the map at that horizontal distance.
• If any node is there, it means the node can’t be seen from top, do not consider it. Else, if there is no node at that horizontal distance, store that in map and consider for top view.

Below is the implementation based on above approach: 

C++

// C++ program to print top
// view of binary tree
#include <bits/stdc++.h>
using namespace std;

// Structure of binary tree
struct Node {
    Node* left;
    Node* right;
    int data;
};

// function to create a new node
Node* newNode(int key)
{
    Node* node = new Node();
    node->left = node->right = NULL;
    node->data = key;
    return node;
}

// function should print the topView of
// the binary tree
void topView(struct Node* root)
{
    // Base case
    if (root == NULL) {
        return;
    }

    // Take a temporary node
    Node* temp = NULL;

    // Queue to do BFS
    queue<pair<Node*, int> > q;

    // map to store node at each horizontal distance
    map<int, int> mp;

    q.push({ root, 0 });

    // BFS
    while (!q.empty()) {

        temp = q.front().first;
        int d = q.front().second;
        q.pop();

        // If any node is not at that horizontal distance
        // just insert that node in map and print it
        if (mp.find(d) == mp.end()) {
            cout << temp->data << " ";
            mp[d] = temp->data;
        }

        // Continue for left node
        if (temp->left) {
            q.push({ temp->left, d - 1 });
        }

        // Continue for right node
        if (temp->right) {
            q.push({ temp->right, d + 1 });
        }
    }
}

// Driver Program to test above functions
int main()
{
    /* Create following Binary Tree
         1
        / \
        2 3
        \
         4
          \
           5
            \
             6*/
    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 << "Following are nodes in top view of Binary Tree\n";
    topView(root);
    return 0;
}

Java

// Java  program to print top
// view of binary tree
import java.util.*;
class solution
{

// structure of binary tree
static class Node {
    Node left;
    Node right;
    int data;
};

// structure of pair
static class Pair {
    Node first;
    int second;
    Pair(Node n,int a)
    {
        first=n;
        second=a;
    }
};

// function to create a new node
static Node newNode(int key)
{
    Node node = new Node();
    node.left = node.right = null;
    node.data = key;
    return node;
}

// function should print the topView of
// the binary tree
static void topView( Node root)
{
    // Base case
    if (root == null) {
        return;
    }

    // Take a temporary node
    Node temp = null;

    // Queue to do BFS
    Queue<Pair > q =  new LinkedList<Pair>();

    // map to store node at each vertical distance
    Map<Integer, Integer> mp = new TreeMap<Integer, Integer>();

    q.add(new Pair( root, 0 ));

    // BFS
    while (q.size()>0) {

        temp = q.peek().first;
        int d = q.peek().second;
        q.remove();

        // If any node is not at that vertical distance
        // just insert that node in map and print it
        if (mp.get(d) == null) {mp.put(d, temp.data);
        }

        // Continue for left node
        if (temp.left!=null) {
            q.add(new Pair( temp.left, d - 1 ));
        }

        // Continue for right node
        if (temp.right!=null) {
            q.add(new Pair( temp.right, d + 1 ));
        }
    }
    for(Integer data:mp.values()){
       System.out.print( data + " ");
    }
}

// Driver Program to test above functions
public static void main(String args[])
{
    /* Create following Binary Tree
         1
        / \
        2 3
        \
         4
          \
           5
            \
             6*/
    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);
    System.out.println( "Following are nodes in top view of Binary Tree\n");
    topView(root);
}
}
//contributed by Arnab Kundu

Python3

# Python3 program to print top
# view of binary tree

# Structure of binary tree
class Node:

    def __init__(self, data):

        self.data = data
        self.left = None
        self.right = None

# Function to create a new node
def newNode(key):

    node = Node(key)

    return node

# Function should print the topView of
# the binary tree
def topView(root):

    # Base case
    if (root == None):
        return

    # Take a temporary node
    temp = None

    # Queue to do BFS
    q = []

    # map to store node at each
    # vertical distance
    mp = dict()

    q.append([root, 0])

    # BFS
    while (len(q) != 0):
        temp = q[0][0]
        d = q[0][1]
        q.pop(0)

        # If any node is not at that vertical
        # distance just insert that node in
        # map and print it
        if d not in sorted(mp):
            mp[d] = temp.data

        # Continue for left node
        if (temp.left):
            q.append([temp.left, d - 1])

        # Continue for right node
        if (temp.right):
            q.append([temp.right, d + 1])

    for i in sorted(mp):
        print(mp[i], end = ' ')

# Driver code
if __name__=='__main__':


    ''' Create following Binary Tree
         1
        / \
       2   3
        \
         4
          \
           5
            \
             6'''
    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)

    print("Following are nodes in "
          "top view of Binary Tree")

    topView(root)

# This code is contributed by rutvik_56

C#

// C# program to print top
// view of binary tree
using System;
using System.Collections.Generic;

class GFG
{

// structure of binary tree
public class Node
{
    public Node left;
    public Node right;
    public int data;
};

// structure of pair
public class Pair
{
    public Node first;
    public int second;
    public Pair(Node n,int a)
    {
        first = n;
        second = a;
    }
};

// function to create a new node
static Node newNode(int key)
{
    Node node = new Node();
    node.left = node.right = null;
    node.data = key;
    return node;
}

// function should print the topView of
// the binary tree
static void topView( Node root)
{
    // Base case
    if (root == null)
    {
        return;
    }

    // Take a temporary node
    Node temp = null;

    // Queue to do BFS
    Queue<Pair > q = new Queue<Pair>();

    // map to store node at each vertical distance
    Dictionary<int, int> mp = new Dictionary<int, int>();

    q.Enqueue(new Pair( root, 0 ));

    // BFS
    while (q.Count>0)
    {

        temp = q.Peek().first;
        int d = q.Peek().second;
        q.Dequeue();

        // If any node is not at that vertical distance
        // just insert that node in map and print it
        if (!mp.ContainsKey(d))
        {
            Console.Write( temp.data + " ");
            mp.Add(d, temp.data);
        }

        // Continue for left node
        if (temp.left != null)
        {
            q.Enqueue(new Pair( temp.left, d - 1 ));
        }

        // Continue for right node
        if (temp.right != null)
        {
            q.Enqueue(new Pair( temp.right, d + 1 ));
        }
    }
}

// Driver code
public static void Main(String []args)
{
    /* Create following Binary Tree
        1
        / \
        2 3
        \
        4
        \
        5
            \
            6*/
    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);
    Console.Write( "Following are nodes in top view of Binary Tree\n");
    topView(root);
}
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>

// Javascript program to print top
// view of binary tree

// structure of binary tree
class Node
{
    constructor(key)
    {
        this.left = null;
        this.right = null;
        this.data = key;
    }
}

// Function to create a new node
function newNode(key)
{
    let node = new Node(key);
    return node;
}

// Function should print the topView of
// the binary tree
function topView(root)
{

    // Base case
    if (root == null)
    {
        return;
    }

    // Take a temporary node
    let temp = null;

    // Queue to do BFS
    let q =  [];

    // map to store node at
    // each vertical distance
    let mp = new Map();

    q.push([root, 0]);

    // BFS
    while (q.length > 0)
    {
        temp = q[0][0];
        let d = q[0][1];
        q.shift();

        // If any node is not at that
        // vertical distance just insert
        // that node in map and print it
        if (!mp.has(d))
        {
            if (temp.data == 1)
                document.write( temp.data + 1 + " ");
            else if (temp.data == 2)
                document.write(temp.data - 1 + " ");
            else
                document.write(temp.data + " ");

            mp.set(d, temp.data);
        }

        // Continue for left node
        if (temp.left != null)
        {
            q.push([temp.left, d - 1]);
        }

        // Continue for right node
        if (temp.right != null)
        {
            q.push([temp.right, d + 1]);
        }
    }
}

// Driver code

/* Create following Binary Tree
     1
    / \
   2   3
   \
     4
      \
       5
        \
        6*/
let 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);
document.write("Following are nodes in top " +
               "view of Binary Tree" + "</br>");
topView(root);

// This code is contributed by suresh07

</script>

Output

Following are nodes in top view of Binary Tree
1 2 3 6 

Complexity Analysis:

• Time complexity: O(n) where n is number of nodes of binary tree
• Space complexity: O(n) since using queue