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 6Approach:
- 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 vertical distance of node from root. Also, maintain a map which should store the node at a particular vertical distance.
- While processing a node, just check if any node is there in the map at that vertical 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 vertical 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 treestruct Node { Node* left; Node* right; int data;};// function to create a new nodeNode* 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 treevoid 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 vartical 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 vertical 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 functionsint 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 treeimport java.util.*;class solution{// structure of binary treestatic class Node { Node left; Node right; int data;};// structure of pairstatic class Pair { Node first; int second; Pair(Node n,int a) { first=n; second=a; }};// function to create a new nodestatic 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 treestatic 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 vartical 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 functionspublic 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 treeclass Node: def __init__(self, data): self.data = data self.left = None self.right = None # Function to create a new nodedef newNode(key): node = Node(key) return node # Function should print the topView of# the binary treedef 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 # vartical 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 codeif __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 treeusing System;using System.Collections.Generic;class GFG{// structure of binary treepublic class Node{ public Node left; public Node right; public int data;};// structure of pairpublic class Pair{ public Node first; public int second; public Pair(Node n,int a) { first = n; second = a; }};// function to create a new nodestatic 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 treestatic 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 vartical 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 codepublic 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 */ |
Output:
Following are nodes in top view of Binary Tree 2 1 3 6
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
In case you wish to attend live classes with industry experts, please refer Geeks Classes Live
