Print nodes in top view of Binary Tree | Set 2
Last Updated :
11 Jul, 2025
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:
C++
// 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;
}
// Java Program to print Top View of Binary Tree
// using hashmap and recursion
import java.util.*;
class GFG {
// Node structure
static class Node {
// Data of the node
int data;
// Reference to left node
Node left;
// Reference to right node
Node right;
};
static class pair {
int data, height;
public pair(int data, int height)
{
this.data = data;
this.height = height;
}
}
// Initialising node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return node;
}
static void printTopViewUtil(Node root, int height,
int hd,
Map<Integer, pair> 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.containsKey(hd)) {
m.put(hd, new pair(root.data, height));
}
else {
pair p = m.get(hd);
if (p.height >= height) {
m.put(hd, new 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);
}
static void printTopView(Node root)
{
// Map to store horizontal distance,
// height and node's data
Map<Integer, pair> m = new TreeMap<>();
printTopViewUtil(root, 0, 0, m);
// Print the node's value stored by
// printTopViewUtil()
for (Map.Entry<Integer, pair> it : m.entrySet()) {
pair p = it.getValue();
System.out.print(p.data + " ");
}
}
// 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.print("Top View : ");
printTopView(root);
}
}
# Python3 Program to prTop View of
# Binary Tree using hash and recursion
from collections import OrderedDict
# A binary tree node
class newNode:
# A constructor to create a
# new Binary tree Node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
self.hd = 2**32
def printTopViewUtil(root, height, hd, m):
# Base Case
if (root == None):
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 hd not in m :
m[hd] = [root.data, height]
else:
p = m[hd]
if p[1] > height:
m[hd] = [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)
def printTopView(root):
# to store horizontal distance,
# height and node's data
m = OrderedDict()
printTopViewUtil(root, 0, 0, m)
# Print the node's value stored
# by printTopViewUtil()
for i in sorted(list(m)):
p = m[i]
print(p[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("Top View : ", end = "")
printTopView(root)
# This code is contributed by SHUBHAMSINGH10
// C# program to print Top View of Binary
// Tree using hashmap and recursion
using System;
using System.Collections.Generic;
class GFG{
// Node structure
class Node
{
// Data of the node
public int data;
// Reference to left node
public Node left;
// Reference to right node
public Node right;
};
class pair
{
public int data, height;
public pair(int data, int height)
{
this.data = data;
this.height = height;
}
}
// Initialising node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return node;
}
static void printTopViewUtil(Node root, int height,
int hd,
SortedDictionary<int, pair> 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.ContainsKey(hd))
{
m[hd] = new pair(root.data, height);
}
else
{
pair p = m[hd];
if (p.height >= height)
{
m[hd] = new 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);
}
static void printTopView(Node root)
{
// Map to store horizontal distance,
// height and node's data
SortedDictionary<int,
pair> m = new SortedDictionary<int,
pair>();
printTopViewUtil(root, 0, 0, m);
// Print the node's value stored by
// printTopViewUtil()
foreach(var it in m.Values)
{
Console.Write(it.data + " ");
}
}
// 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.Write("Top View : ");
printTopView(root);
}
}
// This code is contributed by rutvik_56
<script>
// JavaScript program to print Top View of Binary
// Tree using hashmap and recursion
// Node structure
class Node
{
constructor()
{
// Data of the node
this.data = 0;
// Reference to left node
this.left = null;
// Reference to right node
this.right = null;
}
};
class pair
{
constructor(data, height)
{
this.data = data;
this.height = height;
}
}
// Initialising node
function newNode(data)
{
var node = new Node();
node.data = data;
node.left = null;
node.right = null;
return node;
}
function printTopViewUtil(root, height, hd, 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.has(hd))
{
m.set(hd, new pair(root.data, height));
}
else
{
var p = m.get(hd);
if (p.height >= height)
{
m.set(hd, new 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);
}
function printTopView(root)
{
// Map to store horizontal distance,
// height and node's data
var m = new Map();
printTopViewUtil(root, 0, 0, m);
// Print the node's value stored by
// printTopViewUtil()
for(var it of [...m].sort())
{
document.write(it[1].data + " ");
}
}
// Driver code
var 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("Top View : ");
printTopView(root);
</script>
Complexity Analysis:
- Time complexity: O(n) where n is number of nodes of binary tree
- Auxiliary space: O(n) for call stack
Approach#2: Using deque
We use a level order traversal technique to traverse the tree and maintain a dictionary to store the horizontal distance of each node from the root node. We keep adding nodes to the queue along with their horizontal distance from the root node. Then, we traverse the dictionary and print the values of the nodes with the minimum horizontal distance from the root node.
Algorithm
1. Create an empty dictionary to store the horizontal distance of each node from the root node.
2. Create a queue and enqueue the root node along with its horizontal distance, which is zero.
3. Traverse the tree using a level order traversal technique:
a. Dequeue a node and its horizontal distance from the queue.
b. If the horizontal distance is not present in the dictionary, add the node's value to the dictionary with its horizontal distance.
c. Enqueue the left child of the dequeued node with its horizontal distance decreased by 1.
d. Enqueue the right child of the dequeued node with its horizontal distance increased by 1.
4. Traverse the dictionary and print the values of the nodes with the minimum horizontal distance from the root node.
C++
#include <iostream>
#include <map>
#include <deque>
// Definition of a binary tree node
class Node {
public:
int val;
Node* left;
Node* right;
Node(int value) : val(value), left(nullptr), right(nullptr) {}
};
// Function to print the top view of a binary tree
void topView(Node* root) {
// Map to store horizontal distance and corresponding node's value
std::map<int, int> hdMap;
// Queue for level order traversal with horizontal distance
std::deque<std::pair<Node*, int>> q;
// Enqueue the root with horizontal distance 0
q.push_back(std::make_pair(root, 0));
// Level order traversal to find the top view
while (!q.empty()) {
Node* node = q.front().first;
int hd = q.front().second;
q.pop_front();
// If the horizontal distance is not in the map, add it with the node's value
if (hdMap.find(hd) == hdMap.end()) {
hdMap[hd] = node->val;
}
// Enqueue the left child with a decreased horizontal distance
if (node->left) {
q.push_back(std::make_pair(node->left, hd - 1));
}
// Enqueue the right child with an increased horizontal distance
if (node->right) {
q.push_back(std::make_pair(node->right, hd + 1));
}
}
// Print the top view nodes in order of horizontal distance
for (auto it = hdMap.begin(); it != hdMap.end(); ++it) {
std::cout << it->second << " ";
}
}
int main() {
// Create a sample binary tree
Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);
root->right->left = new Node(6);
root->right->right = new Node(7);
// Print the top view of the binary tree
std::cout << "Top view: ";
topView(root);
return 0;
}
import java.util.*;
class Node {
int val;
Node left;
Node right;
public Node(int val) {
this.val = val;
this.left = null;
this.right = null;
}
}
public class Main {
public static void topView(Node root) {
if (root == null) {
return;
}
Map<Integer, Integer> hdMap = new TreeMap<>();
Queue<Node> queue = new LinkedList<>();
Queue<Integer> hdQueue = new LinkedList<>();
queue.offer(root);
hdQueue.offer(0);
while (!queue.isEmpty()) {
Node currNode = queue.poll();
int currHd = hdQueue.poll();
if (!hdMap.containsKey(currHd)) {
hdMap.put(currHd, currNode.val);
}
if (currNode.left != null) {
queue.offer(currNode.left);
hdQueue.offer(currHd - 1);
}
if (currNode.right != null) {
queue.offer(currNode.right);
hdQueue.offer(currHd + 1);
}
}
for (int hd : hdMap.keySet()) {
System.out.print(hdMap.get(hd) + " ");
}
}
public static void main(String[] args) {
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
System.out.print("Top view: ");
topView(root);
}
}
from collections import deque
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
def top_view(root):
hd_dict = {}
q = deque([(root, 0)])
while q:
node, hd = q.popleft()
if hd not in hd_dict:
hd_dict[hd] = node.val
if node.left:
q.append((node.left, hd-1))
if node.right:
q.append((node.right, hd+1))
for hd in sorted(hd_dict.keys()):
print(hd_dict[hd], end=' ')
# Example usage
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
print("Top view: ", end='')
top_view(root)
using System;
using System.Collections.Generic;
using System.Linq; // Adding System.Linq for LINQ methods
public class Node
{
public int val;
public Node left;
public Node right;
public Node(int val)
{
this.val = val;
left = null;
right = null;
}
}
public class BinaryTree
{
public static void TopView(Node root)
{
Dictionary<int, int> hdDict = new Dictionary<int, int>();
Queue<(Node, int)> queue = new Queue<(Node, int)>();
queue.Enqueue((root, 0));
while (queue.Count > 0)
{
(Node node, int hd) = queue.Dequeue();
if (!hdDict.ContainsKey(hd))
{
hdDict[hd] = node.val;
}
if (node.left != null)
{
queue.Enqueue((node.left, hd - 1));
}
if (node.right != null)
{
queue.Enqueue((node.right, hd + 1));
}
}
foreach (int hd in hdDict.Keys.OrderBy(key => key)) // Using OrderBy from System.Linq
{
Console.Write(hdDict[hd] + " ");
}
}
}
class Program
{
static void Main(string[] args)
{
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
Console.Write("Top view: ");
BinaryTree.TopView(root);
}
}
// Node class definition
class Node {
constructor(val) {
this.val = val;
this.left = null;
this.right = null;
}
}
// Function to find and print the top view of a binary tree
function topView(root) {
const hdMap = {}; // Dictionary to store nodes at each horizontal distance
const queue = [{ node: root, hd: 0 }]; // Queue for level order traversal
while (queue.length !== 0) {
const { node, hd } = queue.shift();
if (!(hd in hdMap)) {
hdMap[hd] = node.val;
}
if (node.left !== null) {
queue.push({ node: node.left, hd: hd - 1 });
}
if (node.right !== null) {
queue.push({ node: node.right, hd: hd + 1 });
}
}
// Print the top view nodes
Object.keys(hdMap)
.sort((a, b) => a - b) // Sorting keys to print nodes in left-to-right order
.forEach(hd => process.stdout.write(hdMap[hd] + ' '));
}
// Example usage
const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
process.stdout.write("Top view: ");
topView(root);
OutputTop view: 4 2 1 3 7
Time Complexity: O(N), where N is the number of nodes in the binary tree, as we need to visit each node once.
Space Complexity: O(N), where N is the maximum number of nodes at any level in the binary tree, as we need to store the horizontal distance of each node from the root node. Additionally, we need to store the nodes in the queue, which can have at most N nodes in the worst case when the tree is a complete binary tree.
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