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:
C++
// C++ Program to print Top View of Binary Tree// using hashmap and recursion#include <bits/stdc++.h>using namespace std;// Node structurestruct 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 nodestruct 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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Java
// Java Program to print Top View of Binary Tree// using hashmap and recursionimport 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); }} |
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Python3
# Python3 Program to prTop View of # Binary Tree using hash and recursionfrom collections import OrderedDict # A binary tree nodeclass 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**32def 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 Coderoot = 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 |
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C#
// C# program to print Top View of Binary// Tree using hashmap and recursionusing System;using System.Collections.Generic;class GFG{ // Node structureclass 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 nodestatic 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 codepublic 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 |
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Output
Top View : 2 1 3 6
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