Print Left View of a Binary Tree
Given a Binary Tree, print left view of it. Left view of a Binary Tree is set of nodes visible when tree is visited from left side.
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Examples:
Input :
1
/ \
2 3
/ \ \
4 5 6
Output : 1 2 4
Input :
1
/ \
2 3
\
4
\
5
\
6
Output :1 2 4 5 6Method-1 (Using Recursion)
The left view contains all nodes that are first nodes in their levels. A simple solution is to do level order traversal and print the first node in every level.
The problem can also be solved using simple recursive traversal. We can keep track of the level of a node by passing a parameter to all recursive calls. The idea is to keep track of the maximum level also. Whenever we see a node whose level is more than maximum level so far, we print the node because this is the first node in its level (Note that we traverse the left subtree before right subtree).
Below is the implementation of the above idea-
C++
// C++ program to print left view of Binary Tree#include <bits/stdc++.h>using namespace std;struct Node{ int data; struct Node *left, *right;};// A utility function to// create a new Binary Tree Nodestruct Node *newNode(int item){ struct Node *temp = (struct Node *)malloc( sizeof(struct Node)); temp->data = item; temp->left = temp->right = NULL; return temp;}// Recursive function to print// left view of a binary tree.void leftViewUtil(struct Node *root, int level, int *max_level){ // Base Case if (root == NULL) return; // If this is the first Node of its level if (*max_level < level) { cout << root->data << " "; *max_level = level; } // Recur for left subtree first, // then right subtree leftViewUtil(root->left, level + 1, max_level); leftViewUtil(root->right, level + 1, max_level); }// A wrapper over leftViewUtil()void leftView(struct Node *root){ int max_level = 0; leftViewUtil(root, 1, &max_level);}// Driver Codeint main(){ Node* root = newNode(10); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(7); root->left->right = newNode(8); root->right->right = newNode(15); root->right->left = newNode(12); root->right->right->left = newNode(14); leftView(root); return 0;} |
C
// C program to print left view of Binary Tree#include <stdio.h>#include <stdlib.h>struct node { int data; struct node *left, *right;};// A utility function to create a new Binary Tree nodestruct node* newNode(int item){ struct node* temp = (struct node*)malloc(sizeof(struct node)); temp->data = item; temp->left = temp->right = NULL; return temp;}// Recursive function to print left view of a binary tree.void leftViewUtil(struct node* root, int level, int* max_level){ // Base Case if (root == NULL) return; // If this is the first node of its level if (*max_level < level) { printf("%d\t", root->data); *max_level = level; } // Recur for left and right subtrees leftViewUtil(root->left, level + 1, max_level); leftViewUtil(root->right, level + 1, max_level);}// A wrapper over leftViewUtil()void leftView(struct node* root){ int max_level = 0; leftViewUtil(root, 1, &max_level);}// Driver codeint main(){ struct node* root = newNode(12); root->left = newNode(10); root->right = newNode(30); root->right->left = newNode(25); root->right->right = newNode(40); leftView(root); return 0;} |
Java
// Java program to print left view of binary tree/* Class containing left and right child of currentnode and key value*/class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; }}/* Class to print the left view */class BinaryTree { Node root; static int max_level = 0; // recursive function to print left view void leftViewUtil(Node node, int level) { // Base Case if (node == null) return; // If this is the first node of its level if (max_level < level) { System.out.print(" " + node.data); max_level = level; } // Recur for left and right subtrees leftViewUtil(node.left, level + 1); leftViewUtil(node.right, level + 1); } // A wrapper over leftViewUtil() void leftView() { leftViewUtil(root, 1); } /* testing for example nodes */ public static void main(String args[]) { /* creating a binary tree and entering the nodes */ BinaryTree tree = new BinaryTree(); tree.root = new Node(12); tree.root.left = new Node(10); tree.root.right = new Node(30); tree.root.right.left = new Node(25); tree.root.right.right = new Node(40); tree.leftView(); }} |
Python
# Python program to print left view of Binary Tree# A binary tree nodeclass Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None# Recursive function print left view of a binary treedef leftViewUtil(root, level, max_level): # Base Case if root is None: return # If this is the first node of its level if (max_level[0] < level): print "% d\t" %(root.data), max_level[0] = level # Recur for left and right subtree leftViewUtil(root.left, level + 1, max_level) leftViewUtil(root.right, level + 1, max_level)# A wrapper over leftViewUtil()def leftView(root): max_level = [0] leftViewUtil(root, 1, max_level)# Driver program to test above functionroot = Node(12)root.left = Node(10)root.right = Node(20)root.right.left = Node(25)root.right.right = Node(40)leftView(root)# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System;// C# program to print left view of binary tree/* Class containing left and right child of currentnode and key value*/public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}/* Class to print the left view */public class BinaryTree { public Node root; public static int max_level = 0; // recursive function to print left view public virtual void leftViewUtil(Node node, int level) { // Base Case if (node == null) { return; } // If this is the first node of its level if (max_level < level) { Console.Write(" " + node.data); max_level = level; } // Recur for left and right subtrees leftViewUtil(node.left, level + 1); leftViewUtil(node.right, level + 1); } // A wrapper over leftViewUtil() public virtual void leftView() { leftViewUtil(root, 1); } /* testing for example nodes */ public static void Main(string[] args) { /* creating a binary tree and entering the nodes */ BinaryTree tree = new BinaryTree(); tree.root = new Node(12); tree.root.left = new Node(10); tree.root.right = new Node(30); tree.root.right.left = new Node(25); tree.root.right.right = new Node(40); tree.leftView(); }}// This code is contributed by Shrikant13 |
Javascript
<script> // Javascript program to print left view// of binary tree// Class containing left and right// child of current node and key valueclass Node{ constructor(item) { this.data = item; this.left = null; this.right = null; }}// Class to print the left viewvar root ;var max_level = 0;// Recursive function to print left viewfunction leftViewUtil(node, level){ // Base Case if (node == null) { return; } // If this is the first node of its level if (max_level < level) { document.write(" " + node.data); max_level = level; } // Recur for left and right subtrees leftViewUtil(node.left, level + 1); leftViewUtil(node.right, level + 1);}// A wrapper over leftViewUtil()function leftView(){ leftViewUtil(root, 1);}// Driver code// Testing for example nodes// Creating a binary tree and// entering the nodesroot = new Node(12);root.left = new Node(10);root.right = new Node(30);root.right.left = new Node(25);root.right.right = new Node(40);leftView();// This code is contributed by rrrtnx</script> |
Output
12 10 25
Time Complexity: The function does a simple traversal of the tree, so the complexity is O(n).
Auxiliary Space: O(n), due to the stack space during recursive call.
Method-2 (Using Queue):
In this method, level order traversal based solution is discussed. If we observe carefully, we will see that our main task is to print the left most node of every level. So, we will do a level order traversal on the tree and print the leftmost node at every level. Below is the implementation of above approach:
C++
// C++ program to print left view of// Binary Tree#include<bits/stdc++.h>using namespace std;// A Binary Tree Nodestruct Node{ int data; struct Node *left, *right;};// Utility function to create a new tree nodeNode* newNode(int data){ Node *temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp;}// function to print left view of// binary treevoid printLeftView(Node* root){ if (!root) return; queue<Node*> q; q.push(root); while (!q.empty()) { // number of nodes at current level int n = q.size(); // Traverse all nodes of current level for(int i = 1; i <= n; i++) { Node* temp = q.front(); q.pop(); // Print the left most element // at the level if (i == 1) cout<<temp->data<<" "; // Add left node to queue if (temp->left != NULL) q.push(temp->left); // Add right node to queue if (temp->right != NULL) q.push(temp->right); } }} // Driver codeint main(){ // Let's construct the tree as // shown in example Node* root = newNode(10); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(7); root->left->right = newNode(8); root->right->right = newNode(15); root->right->left = newNode(12); root->right->right->left = newNode(14); printLeftView(root);}// This code is contributed by// Manne SreeCharan |
Java
// Java program to print left view of Binary// Treeimport java.util.*;public class PrintRightView { // Binary tree node private static class Node { int data; Node left, right; public Node(int data) { this.data = data; this.left = null; this.right = null; } } // function to print left view of binary tree private static void printLeftView(Node root) { if (root == null) return; Queue<Node> queue = new LinkedList<>(); queue.add(root); while (!queue.isEmpty()) { // number of nodes at current level int n = queue.size(); // Traverse all nodes of current level for (int i = 1; i <= n; i++) { Node temp = queue.poll(); // Print the left most element at // the level if (i == 1) System.out.print(temp.data + " "); // Add left node to queue if (temp.left != null) queue.add(temp.left); // Add right node to queue if (temp.right != null) queue.add(temp.right); } } } // Driver code public static void main(String[] args) { // construct binary tree as shown in // above diagram Node root = new Node(10); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(7); root.left.right = new Node(8); root.right.right = new Node(15); root.right.left = new Node(12); root.right.right.left = new Node(14); printLeftView(root); }}// This code is contributed by// Manne SreeCharan |
Python
# Python3 program to print left view of# Binary Tree# Binary Tree Node""" utility that allocates a newNodewith 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 to print left view of# binary treedef printRightView(root): if (not root): return q = [] q.append(root) while (len(q)): # number of nodes at current level n = len(q) # Traverse all nodes of current level for i in range(1, n + 1): temp = q[0] q.pop(0) # Print the left most element # at the level if (i == 1): print(temp.data, end=" ") # Add left node to queue if (temp.left != None): q.append(temp.left) # Add right node to queue if (temp.right != None): q.append(temp.right)# Driver Codeif __name__ == '__main__': root = newNode(10) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(7) root.left.right = newNode(8) root.right.right = newNode(15) root.right.left = newNode(12) root.right.right.left = newNode(14) printRightView(root)# This code is contributed by# Manne SreeCharan |
C#
// C# program to print left view// of Binary Treeusing System;using System.Collections.Generic;public class PrintRightView { // Binary tree node private class Node { public int data; public Node left, right; public Node(int data) { this.data = data; this.left = null; this.right = null; } } // function to print left view of binary tree private static void printRightView(Node root) { if (root == null) return; Queue<Node> queue = new Queue<Node>(); queue.Enqueue(root); while (queue.Count != 0) { // number of nodes at current level int n = queue.Count; // Traverse all nodes of current level for (int i = 1; i <= n; i++) { Node temp = queue.Dequeue(); // Print the left most element at // the level if (i == n) Console.Write(temp.data + " "); // Add left node to queue if (temp.left != null) queue.Enqueue(temp.left); // Add right node to queue if (temp.right != null) queue.Enqueue(temp.right); } } } // Driver code public static void Main(String[] args) { // construct binary tree as shown in // above diagram Node root = new Node(10); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(7); root.left.right = new Node(8); root.right.right = new Node(15); root.right.left = new Node(12); root.right.right.left = new Node(14); printRightView(root); }}// This code is contributed Manne SreeCharan |
Output
10 2 7 14
Time Complexity: O(n), where n is the number of nodes in the binary tree.
This article is contributed by Ramsai Chinthamani, Manne SreeCharan Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
