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 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
// 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
# 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#
// 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 |
Javascript
<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> |
Top View : 2 1 3 6
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.
Python3
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) |
Java
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); } } |
Top 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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