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Print Nodes in Top View of Binary Tree

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  • Difficulty Level : Medium
  • Last Updated : 04 Sep, 2022
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The 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. The horizontal distance of the left child of a node x is equal to a horizontal distance of x minus 1, and that of a right child is the horizontal distance of x plus 1. 

       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

The idea is to do something similar to vertical Order Traversal. Like vertical Order Traversal, we need to put nodes of the 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 the same horizontal distance below it. Hashing is used to check if a node at a given horizontal distance is seen or not. 

Below is the implementation of the above approach:

C++14




// 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 hd;
    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(Node* root)
{
    if (root == NULL)
        return;
    queue<Node*> q;
    map<int, int> m;
    int hd = 0;
    root->hd = hd;
 
    // push node and horizontal distance to queue
    q.push(root);
 
    cout << "The top view of the tree is : \n";
 
    while (q.size()) {
        hd = root->hd;
 
        // count function returns 1 if the container
        // contains an element whose key is equivalent
        // to hd, or returns zero otherwise.
        if (m.count(hd) == 0)
            m[hd] = root->data;
        if (root->left) {
            root->left->hd = hd - 1;
            q.push(root->left);
        }
        if (root->right) {
            root->right->hd = hd + 1;
            q.push(root->right);
        }
        q.pop();
        root = q.front();
    }
 
    for (auto i = m.begin(); i != m.end(); i++) {
        cout << i->second << " ";
    }
}
 
// Driver code
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;
}
/* This code is contributed by Niteesh Kumar */


Java




// Java program to print top
// view of binary tree
import java.util.LinkedList;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Queue;
import java.util.TreeMap;
 
// class to create a node
class Node {
    int data;
    Node left, right;
 
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}
 
// class of binary tree
class BinaryTree {
    Node root;
 
    public BinaryTree() { root = null; }
 
    // function should print the topView of
    // the binary tree
    private void TopView(Node root)
    {
        class QueueObj {
            Node node;
            int hd;
 
            QueueObj(Node node, int hd)
            {
                this.node = node;
                this.hd = hd;
            }
        }
        Queue<QueueObj> q = new LinkedList<QueueObj>();
        Map<Integer, Node> topViewMap
            = new TreeMap<Integer, Node>();
 
        if (root == null) {
            return;
        }
        else {
            q.add(new QueueObj(root, 0));
        }
 
        System.out.println(
            "The top view of the tree is : ");
 
        // count function returns 1 if the container
        // contains an element whose key is equivalent
        // to hd, or returns zero otherwise.
        while (!q.isEmpty()) {
            QueueObj tmpNode = q.poll();
            if (!topViewMap.containsKey(tmpNode.hd)) {
                topViewMap.put(tmpNode.hd, tmpNode.node);
            }
 
            if (tmpNode.node.left != null) {
                q.add(new QueueObj(tmpNode.node.left,
                                   tmpNode.hd - 1));
            }
            if (tmpNode.node.right != null) {
                q.add(new QueueObj(tmpNode.node.right,
                                   tmpNode.hd + 1));
            }
        }
        for (Map.Entry<Integer, Node> entry :
             topViewMap.entrySet()) {
            System.out.print(entry.getValue().data + " ");
        }
    }
 
    // Driver Program to test above functions
    public static void main(String[] args)
    {
        /* Create following Binary Tree
        1
       / \
      2   3
       \
         4
          \
           5
             \
              6
   */
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.right = new Node(4);
        tree.root.left.right.right = new Node(5);
        tree.root.left.right.right.right = new Node(6);
        System.out.println(
            "Following are nodes in top view of Binary Tree");
        tree.TopView(tree.root);
    }
}


Python3




# Python3 program to print top
# view of binary tree
 
# Binary Tree Node
""" utility that allocates a newNode
with the given key """
 
 
class newNode:
 
    # Construct to create a newNode
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
        self.hd = 0
 
# function should print the topView
# of the binary tree
 
 
def topview(root):
 
    if(root == None):
        return
    q = []
    m = dict()
    hd = 0
    root.hd = hd
 
    # push node and horizontal
    # distance to queue
    q.append(root)
 
    while(len(q)):
        root = q[0]
        hd = root.hd
 
        # count function returns 1 if the
        # container contains an element
        # whose key is equivalent to hd,
        # or returns zero otherwise.
        if hd not in m:
            m[hd] = root.data
        if(root.left):
            root.left.hd = hd - 1
            q.append(root.left)
 
        if(root.right):
            root.right.hd = hd + 1
            q.append(root.right)
 
        q.pop(0)
    for i in sorted(m):
        print(m[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
# Shubham Singh(SHUBHAMSINGH10)


C#




// C# program to print top
// view of binary tree
using System;
using System.Collections;
using System.Collections.Generic;
 
// Class to create a node
class Node {
    public int data;
    public Node left, right;
 
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
};
 
class QueueObj {
    public Node node;
    public int hd;
 
    public QueueObj(Node node, int hd)
    {
        this.node = node;
        this.hd = hd;
    }
};
 
// Class of binary tree
class BinaryTree {
 
    Node root;
 
    public BinaryTree() { root = null; }
 
    // function should print the topView of
    // the binary tree
    void TopView(Node root)
    {
        Queue q = new Queue();
        SortedDictionary<int, Node> topViewMap
            = new SortedDictionary<int, Node>();
 
        if (root == null) {
            return;
        }
        else {
            q.Enqueue(new QueueObj(root, 0));
        }
 
        // count function returns 1 if the container
        // contains an element whose key is equivalent
        // to hd, or returns zero otherwise.
        while (q.Count != 0) {
            QueueObj tmpNode = (QueueObj)q.Dequeue();
 
            if (!topViewMap.ContainsKey(tmpNode.hd)) {
                topViewMap[tmpNode.hd] = tmpNode.node;
            }
 
            if (tmpNode.node.left != null) {
                q.Enqueue(new QueueObj(tmpNode.node.left,
                                       tmpNode.hd - 1));
            }
            if (tmpNode.node.right != null) {
                q.Enqueue(new QueueObj(tmpNode.node.right,
                                       tmpNode.hd + 1));
            }
        }
 
        foreach(var entry in topViewMap.Values)
        {
            Console.Write(entry.data);
        }
    }
 
    // Driver code
    public static void Main(string[] args)
    {
 
        /* Create following Binary Tree
            1
           / \
          2   3
           \
            4
             \
              5
               \
                6*/
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.right = new Node(4);
        tree.root.left.right.right = new Node(5);
        tree.root.left.right.right.right = new Node(6);
 
        Console.WriteLine("Following are nodes "
                          + "in top view of Binary Tree");
 
        tree.TopView(tree.root);
    }
}
 
// This code is contributed by rutvik_56


Javascript




<script>
// Javascript program to print top
// view of binary tree
 
class Node
{
    constructor(data)
    {
        this.data=data;
        this.left = this.right = null;
        this.hd = 0;
    }
}
 
// Driver Code
function topview(root)
{
    if(root == null)
        return;
    let q = [];
    let m = new Map();
    let hd = 0;
    root.hd = hd;
    q.push(root);
     
    while(q.length!=0)
    {
        root = q[0];
        hd = root.hd;
        if(!m.has(hd))
            m.set(hd,root.data);
        if(root.left)
        {
            root.left.hd = hd - 1;
            q.push(root.left);
        }
        if(root.right)
        {
            root.right.hd = hd + 1;
            q.push(root.right);
        }
        q.shift()
    }
     
    let arr = Array.from(m);
    arr.sort(function(a,b){return a[0]-b[0];})
     
    for (let [key, value] of arr.values())
    {
        document.write(value+" ");
    }
}
 
let root = new Node(1)
root.left = new Node(2)
root.right = new Node(3)
root.left.right = new Node(4)
root.left.right.right = new Node(5)
root.left.right.right.right = new Node(6)
document.write("Following are nodes in top",
      "view of Binary Tree<br>")
topview(root)
 
// This code is contributed by avanitrachhadiya2155
</script>


Output

Following are nodes in top view of Binary Tree
The top view of the tree is : 
2 1 3 6 

Time complexity: O(n*log(n)), where n is the number of nodes in the given tree.
Auxiliary Space : O(n), As we store nodes in the map and queue.

HashMap Approach : Improvement over the TreeMap approach

Since we need the horizontal distance in sorted order TreeMap was used in the above solution; but instead, a minimum and maximum horizontal distance variable can be maintained for each iteration.

After that traverse from minimum to maximum while printing the first acquired node during the traversal of the tree that was stored in the map.

Since, each horizontal distance from minimum to maximum is guaranteed to have at least one node in the map

Java




import java.util.*;
 
class GFG {
      // Structure of binary tree
    static class Node {
        Node left;
        Node right;
        int data;
       
          Node(int data) {
            this.left = this.right = null;
              this.data = data;
        }
    }
   
      //Queue Object Structure
    static class QueueObj {
      Node node;
      int hd;
 
      QueueObj(Node node, int hd)
      {
        this.node = node;
        this.hd = hd;
      }
    }
 
    static void topView(Node root)
    {
        if (root == null)
            return;
 
        Queue<QueueObj> q = new LinkedList<>();
        Map<Integer, Integer> map = new HashMap<>();
        int min = 0;
        int max = 0;
        //Level Order Traversal
        q.add(new QueueObj(root, 0));
        while (!q.isEmpty()) {
            QueueObj curr = q.poll();
             
              //only include in map if this is the
              //first node of this specific
              //horizontal distance
            if (!map.containsKey(curr.hd)) {
                map.put(curr.hd, curr.node.data);
            }
             
               
            if (curr.node.left != null) {
                  //min can be found only in left side due to "-1"
                  //minimum horizontal distance of any node from root
                min = Math.min(min, curr.hd - 1);
                q.add(
                    new QueueObj(curr.node.left, curr.hd - 1));
            }
 
            if (curr.node.right != null) {
                  //max can be found only in right side due to "+1"
                  //maximum horizontal distance of any node from root
                max = Math.max(max, curr.hd + 1);
                q.add(
                    new QueueObj(curr.node.right, curr.hd + 1));
            }
        }
         
          //traversal of (horizontal distance from root)
          //minimum to maximum
        for (; min <= max; min++) {
            System.out.print(map.get(min)+" ");
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.right = new Node(4);
        root.left.right.right = new Node(5);
        root.left.right.right.right = new Node(6);
        System.out.println("Following are nodes in"
                           + " top view of Binary Tree");
        topView(root);
    }
}
//Code Contributed by Animesh Singh


Output

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

Time Complexity: O(N), Since we only perform level-order traversal and print some part of the N nodes which at max will be 2N in case of skew tree

Auxiliary Space: O(N), Since we store the nodes in the map and queue.

Another approach: This approach does not require a queue. Here we use the two variables, one for the vertical distance of the current node from the root and another for the depth of the current node from the root. We use the vertical distance for indexing. If one node with the same vertical distance comes again, we check if the depth of the new node is lower or higher with respect to the current node with the same vertical distance in the map. If depth of new node is lower, then we replace it.

Below is the implementation of the above approach:

C++




#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 to fill the map
void fillMap(Node* root, int d, int l,
             map<int, pair<int, int> >& m)
{
    if (root == NULL)
        return;
 
    if (m.count(d) == 0) {
        m[d] = make_pair(root->data, l);
    }
    else if (m[d].second > l) {
        m[d] = make_pair(root->data, l);
    }
 
    fillMap(root->left, d - 1, l + 1, m);
    fillMap(root->right, d + 1, l + 1, m);
}
 
// function should print the topView of
// the binary tree
void topView(struct Node* root)
{
 
    // map to store the pair of node value and its level
    // with respect to the vertical distance from root.
    map<int, pair<int, int> > m;
 
    // fillmap(root,vectical_distance_from_root,level_of_node,map)
    fillMap(root, 0, 0, m);
 
    for (auto it = m.begin(); it != m.end(); it++) {
        cout << it->second.first << " ";
    }
}
// Driver Program to test above functions
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 << "Following are nodes in top view of Binary "
            "Tree\n";
    topView(root);
    return 0;
}
/* This code is contributed by Akash Debnath */


Java




// Java program to print top
// view of binary tree
import java.util.*;
 
class GFG {
 
    // Structure of binary tree
    static class Node {
        Node left;
        Node right;
        int data;
    }
 
    static class pair {
        int first, second;
 
        pair() {}
        pair(int i, int j)
        {
            first = i;
            second = j;
        }
    }
 
    // map to store the pair of node value and
    // its level with respect to the vertical
    // distance from root.
    static TreeMap<Integer, pair> m = new TreeMap<>();
 
    // 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 to fill the map
    static void fillMap(Node root, int d, int l)
    {
        if (root == null)
            return;
 
        if (m.get(d) == null) {
            m.put(d, new pair(root.data, l));
        }
        else if (m.get(d).second > l) {
            m.put(d, new pair(root.data, l));
        }
 
        fillMap(root.left, d - 1, l + 1);
        fillMap(root.right, d + 1, l + 1);
    }
 
    // function should print the topView of
    // the binary tree
    static void topView(Node root)
    {
        fillMap(root, 0, 0);
 
        for (Map.Entry<Integer, pair> entry :
             m.entrySet()) {
            System.out.print(entry.getValue().first + " ");
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        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");
        topView(root);
    }
}
 
// This code is contributed by Arnab Kundu


Python3




# Binary Tree Node
""" utility that allocates a newNode
with the given key """
 
 
class newNode:
 
    # Construct to create a newNode
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
# function to fill the map
 
 
def fillMap(root, d, l, m):
    if(root == None):
        return
 
    if d not in m:
        m[d] = [root.data, l]
    elif(m[d][1] > l):
        m[d] = [root.data, l]
    fillMap(root.left, d - 1, l + 1, m)
    fillMap(root.right, d + 1, l + 1, m)
 
# function should print the topView of
# the binary tree
 
 
def topView(root):
 
    # map to store the pair of node value and its level
    # with respect to the vertical distance from root.
    m = {}
 
    fillMap(root, 0, 0, m)
    for it in sorted(m.keys()):
        print(m[it][0], end=" ")
 
 
# Driver Code
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 SHUBHAMSINGH10


C#




// C# program to print top
// view of binary tree
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG {
 
    // Structure of binary tree
    class Node {
        public Node left;
        public Node right;
        public int data;
    }
 
    class pair {
        public int first, second;
 
        public pair(int i, int j)
        {
            first = i;
            second = j;
        }
    }
 
    // map to store the pair of node value and
    // its level with respect to the vertical
    // distance from root.
    static SortedDictionary<int, pair> m
        = new SortedDictionary<int, pair>();
 
    // 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 to fill the map
    static void fillMap(Node root, int d, int l)
    {
        if (root == null)
            return;
 
        if (!m.ContainsKey(d)) {
            m[d] = new pair(root.data, l);
        }
        else if (m[d].second > l) {
            m[d] = new pair(root.data, l);
        }
 
        fillMap(root.left, d - 1, l + 1);
        fillMap(root.right, d + 1, l + 1);
    }
 
    // function should print the topView of
    // the binary tree
    static void topView(Node root)
    {
        fillMap(root, 0, 0);
 
        foreach(pair entry in m.Values)
        {
            Console.Write(entry.first + " ");
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        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.WriteLine("Following are nodes in"
                          + " top view of Binary Tree");
        topView(root);
    }
}
 
// This code is contributed by pratham76


Javascript




<script>
// Javascript program to print top
// view of binary tree
 
// Structure of binary tree
class Node
{
    constructor()
    {
        this.data = 0;
        this.left = this.right = null;
    }
}
 
class pair
{
    constructor(i, j)
    {
        this.first = i;
        this.second = j;
    }
}
 
// map to store the pair of node value and
    // its level with respect to the vertical
    // distance from root.
let m = new Map();
     
// function to create a new node   
function newNode(key)
{
    let node = new Node();
        node.left = node.right = null;
        node.data = key;
        return node;
}
 
// function to fill the map
function fillMap(root, d, l)
{
    if (root == null)
            return;
  
        if (m.get(d) == null) {
            m.set(d, new pair(root.data, l));
        }
        else if (m.get(d).second > l) {
            m.set(d, new pair(root.data, l));
        }
  
        fillMap(root.left, d - 1, l + 1);
        fillMap(root.right, d + 1, l + 1);
}
 
// function should print the topView of
    // the binary tree
function topView(root)
{
    fillMap(root, 0, 0);
      
    let arr=Array.from(m.keys());
     
    arr.sort(function(a,b){return a-b;});
        for (let key of arr.values()) {
            document.write(m.get(key).first + " ");
        }
}
 
// Driver Code
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 rag2127
</script>


Output

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

Time complexity : O(n*logn), where n is the number of nodes in the given binary tree with each insertion operation in Map requiring O(log2n) complexity.
Auxiliary Space : O(n)

Another approach:

  1. This approach is based on the level order traversal. We’ll keep a record of the current max so far left, right horizontal distances from the root.
  2. And if we found less distance (or greater in magnitude) then max left so far distance then update it and also push data on this node to a stack (stack is used because in level order traversal the left nodes will appear in reverse order), or if we found greater distance then max right so far distance then update it and also push data on this node to a vector.
  3. In this approach, no map is used.

Below is the implementation of the above approach:

C++14




// C++ Program to print Top View of a binary Tree
 
#include <iostream>
#include <queue>
#include <stack>
using namespace std;
 
// class for Tree node
class Node {
public:
    Node *left, *right;
    int data;
    Node() { left = right = 0; }
    Node(int data)
    {
        left = right = 0;
        this->data = data;
    }
};
 
/*
          1
         / \
        2   3
         \
          4
           \
            5
             \
              6
    Top view of the above binary tree is
    2 1 3 6
*/
 
// class for Tree
class Tree {
public:
    Node* root;
    Tree() { root = 0; }
 
    void topView()
    {
        // queue for holding nodes and their horizontal
        // distance from the root node
        queue<pair<Node*, int> > q;
 
        // pushing root node with distance 0
        q.push(make_pair(root, 0));
 
        // hd is current node's horizontal distance from
        // root node l is current left min horizontal
        // distance (or max in magnitude) so far from the
        // root node r is current right max horizontal
        // distance so far from the root node
 
        int hd = 0, l = 0, r = 0;
 
        // stack is for holding left node's data because
        // they will appear in reverse order that is why
        // using stack
        stack<int> left;
 
        // vector is for holding right node's data
        vector<int> right;
 
        Node* node;
 
        while (q.size()) {
 
            node = q.front().first;
            hd = q.front().second;
 
            if (hd < l) {
                left.push(node->data);
                l = hd;
            }
            else if (hd > r) {
                right.push_back(node->data);
                r = hd;
            }
 
            if (node->left) {
                q.push(make_pair(node->left, hd - 1));
            }
            if (node->right) {
                q.push(make_pair(node->right, hd + 1));
            }
 
            q.pop();
        }
        // printing the left node's data in reverse order
        while (left.size()) {
            cout << left.top() << " ";
            left.pop();
        }
 
        // then printing the root node's data
        cout << root->data << " ";
 
        // finally printing the right node's data
        for (auto x : right) {
            cout << x << " ";
        }
    }
};
 
// Driver code
int main()
{
    // Tree object
    Tree t;
    t.root = new Node(1);
    t.root->left = new Node(2);
    t.root->right = new Node(3);
    t.root->left->right = new Node(4);
    t.root->left->right->right = new Node(5);
    t.root->left->right->right->right = new Node(6);
    t.topView();
    cout << endl;
    return 0;
}


Java




// Java program to print top
// view of binary tree
import java.util.*;
 
class GFG {
 
    // Structure of binary tree
    static class Node {
        Node left;
        Node right;
        int data;
    }
/*
          1
         / \
        2   3
         \
          4
           \
            5
             \
              6
    Top view of the above binary tree is
    2 1 3 6
*/
    static class pair {
        Node node;
        int hd;
 
        pair() {}
        pair(Node node, int hd)
        {
            this.node = node;
            this.hd = hd;
        }
    }
 
    // 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)
    {
        // queue for holding nodes and their horizontal
        // distance from the root node
        Queue<pair> q = new LinkedList<>();
 
        // pushing root node with distance 0
        q.add(new pair(root, 0));
 
        // hd is current node's horizontal distance from
        // root node l is current left min horizontal
        // distance (or max in magnitude) so far from the
        // root node r is current right max horizontal
        // distance so far from the root node
 
        int hd = 0, l = 0, r = 0;
 
        // stack is for holding left node's data because
        // they will appear in reverse order that is why
        // using stack
 
        Stack<Integer> left = new Stack<>();
 
        // ArrayList is for holding right node's data
        ArrayList<Integer> right = new ArrayList<>();
        Node node = null;
 
        while (!q.isEmpty()) {
            node = q.peek().node;
            hd = q.peek().hd;
 
            if (hd < l) {
                left.push(node.data);
                l = hd;
            }
 
            if (hd > r) {
                right.add(node.data);
                r = hd;
            }
 
            if (node.left != null) {
                q.add(new pair(node.left, hd - 1));
            }
            if (node.right != null) {
                q.add(new pair(node.right, hd + 1));
            }
 
            q.poll();
        }
 
        // printing the left node's data in reverse order
        while (!left.isEmpty()) {
            System.out.print(left.peek() + " ");
            left.pop();
        }
 
        // then printing the root node's data
        System.out.print(root.data + " ");
 
        // finally printing the right node's data
        for (int d : right) {
            System.out.print(d + " ");
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        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");
        topView(root);
    }
}
 
// This code is contributed by Snigdha Patil


Python3




# Python Program to print Top View of a binary Tree
 
# Class of tree node
 
 
class Node:
    def __init__(self, data):
        self.left = None
        self.right = None
        self.data = data
 
 
'''
          1
         / \
        2   3
         \
          4
          \
            5
             \
              6
    Top view of the above binary tree is
    2 1 3 6
'''
 
# class for Tree
 
 
class Tree:
    def __init__(self):
        self.root = None
 
    def topView(self):
 
        # queue for holding nodes and their horizontal
        # distance from the root node
        q = []
 
        # pushing root node with distance 0
        q.append((self.root, 0))
 
        # hd is current node's horizontal distance from
        #  root node l is current left min horizontal
        #  distance (or max in magnitude) so far from the
        #  root node r is current right max horizontal
        # distance so far from the root node
 
        hd = 0
        l = 0
        r = 0
 
        # stack is for holding left node's data because
        # they will appear in reverse order that is why
        # using stack
        left = []
 
        # list is for holding right node's data
        right = []
 
        while len(q) > 0:
            node, hd = q[0]
 
            if hd < l:
                left.append(node.data)
                l = hd
 
            elif hd > r:
                right.append(node.data)
                r = hd
 
            if node.left != None:
                q.append((node.left, hd-1))
 
            if node.right != None:
                q.append((node.right, hd+1))
            q.pop(0)
 
        # printing the left node's data in reverse order
        while len(left) > 0:
            x = left.pop()
            print(x, end=" ")
 
        # then printing the root node's data
        print(self.root.data, end=" ")
 
        # finally printing the right node's data
        for x in right:
            print(x, end=" ")
 
 
# Driver code
if __name__ == '__main__':
    # Tree object
    t = Tree()
    t.root = Node(1)
    t.root.left = Node(2)
    t.root.right = Node(3)
    t.root.left.right = Node(4)
    t.root.left.right.right = Node(5)
    t.root.left.right.right.right = Node(6)
    t.topView()
 
    print()
 
# This code is contributed by Tapesh(tapeshdua420)


Output

2 1 3 6 

Time Complexity: O(n), where n is the number of nodes in the given binary tree.
Auxiliary Space: O(n)


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