Given a tree, our task is to check whether its left view is sorted or not. If it is then return true else false.
Examples:
Input:
Output: true
Explanation:
The left view for the tree would be 10, 20, 50 which is in sorted order.
Approach:
To solve the problem mentioned above we have to perform level order traversal on the tree and look for the very first node of each level. Then initialize a variable to check whether its left view is sorted or not. If it is not sorted, then we can break the loop and print false else loop goes on and at last print true.
Below is the implementation of the above approach:
C++
// C++ implementation to Check if the Left// View of the given tree is Sorted or not#include <bits/stdc++.h>using namespace std;// Binary Tree Nodestruct node { int val; struct node *right, *left;};// Utility function to create a new nodestruct node* newnode(int key){ struct node* temp = new node; temp->val = key; temp->right = NULL; temp->left = NULL; return temp;}// Function to find left view// and check if it is sortedvoid func(node* root){ // queue to hold values queue<node*> q; // variable to check whether // level order is sorted or not bool t = true; q.push(root); int i = -1, j = -1, k = -1; // Iterate until the queue is empty while (!q.empty()) { int h = q.size(); // Traverse every level in tree while (h > 0) { root = q.front(); // variable for initial level if (i == -1) { j = root->val; } // checking values are sorted or not if (i == -2) { if (j <= root->val) { j = root->val; i = -3; } else { t = false; break; } } // Push left value if it is not null if (root->left != NULL) { q.push(root->left); } // Push right value if it is not null if (root->right != NULL) { q.push(root->right); } h = h - 1; // Pop out the values from queue q.pop(); } i = -2; // Check if the value are not // sorted then break the loop if (t == false) { break; } } if (t) cout << "true" << endl; else cout << "false" << endl;}// Driver codeint main(){ struct node* root = newnode(10); root->right = newnode(50); root->right->right = newnode(15); root->left = newnode(20); root->left->left = newnode(50); root->left->right = newnode(23); root->right->left = newnode(10); func(root); return 0;} |
Java
// Java implementation to check if the left// view of the given tree is sorted or notimport java.util.*;class GFG{// Binary Tree Nodestatic class node { int val; node right, left;};// Utility function to create a new nodestatic node newnode(int key){ node temp = new node(); temp.val = key; temp.right = null; temp.left = null; return temp;}// Function to find left view// and check if it is sortedstatic void func(node root){ // Queue to hold values Queue<node> q = new LinkedList<>(); // Variable to check whether // level order is sorted or not boolean t = true; q.add(root); int i = -1, j = -1; // Iterate until the queue is empty while (!q.isEmpty()) { int h = q.size(); // Traverse every level in tree while (h > 0) { root = q.peek(); // Variable for initial level if (i == -1) { j = root.val; } // Checking values are sorted or not if (i == -2) { if (j <= root.val) { j = root.val; i = -3; } else { t = false; break; } } // Push left value if it is not null if (root.left != null) { q.add(root.left); } // Push right value if it is not null if (root.right != null) { q.add(root.right); } h = h - 1; // Pop out the values from queue q.remove(); } i = -2; // Check if the value are not // sorted then break the loop if (t == false) { break; } } if (t) System.out.print("true" + "\n"); else System.out.print("false" + "\n");}// Driver codepublic static void main(String[] args){ node root = newnode(10); root.right = newnode(50); root.right.right = newnode(15); root.left = newnode(20); root.left.left = newnode(50); root.left.right = newnode(23); root.right.left = newnode(10); func(root);}}// This code is contributed by gauravrajput1 |
Python3
# Python implementation for the above approach# Binary Tree Nodeclass Node: def __init__(self, key): self.left = None self.right = None self.val = key# Utility function to create a new nodedef new_node(key): temp = Node(key) return temp# Function to find left view# and check if it is sorteddef func(root): # Queue to hold values q = [] # Variable to check whether # level order is sorted or not t = True q.append(root) i = -1 j = -1 # Iterate until the queue is empty while len(q) > 0: h = len(q) # Traverse every level in tree while h > 0: root = q[0] # Variable for initial level if i == -1: j = root.val # Checking values are sorted or not if i == -2: if j <= root.val: j = root.val i = -3 else: t = False break # Push left value if it is not null if root.left is not None: q.append(root.left) # Push right value if it is not null if root.right is not None: q.append(root.right) h = h - 1 # Pop out the values from queue q.pop(0) i = -2 # Check if the value are not # sorted then break the loop if t == False: break if t: print("true") else: print("false")# Driver coderoot = new_node(10)root.right = new_node(50)root.right.right = new_node(15)root.left = new_node(20)root.left.left = new_node(50)root.left.right = new_node(23)root.right.left = new_node(10)func(root)# This code is contributed by Potta Lokesh |
C#
// C# implementation to check if the left// view of the given tree is sorted or notusing System;using System.Collections.Generic;class GFG{// Binary Tree Nodeclass node { public int val; public node right, left;};// Utility function to create a new nodestatic node newnode(int key){ node temp = new node(); temp.val = key; temp.right = null; temp.left = null; return temp;}// Function to find left view// and check if it is sortedstatic void func(node root){ // Queue to hold values Queue<node> q = new Queue<node>(); // Variable to check whether // level order is sorted or not bool t = true; q.Enqueue(root); int i = -1, j = -1; // Iterate until the queue is empty while (q.Count != 0) { int h = q.Count; // Traverse every level in tree while (h > 0) { root = q.Peek(); // Variable for initial level if (i == -1) { j = root.val; } // Checking values are sorted or not if (i == -2) { if (j <= root.val) { j = root.val; i = -3; } else { t = false; break; } } // Push left value if it is not null if (root.left != null) { q.Enqueue(root.left); } // Push right value if it is not null if (root.right != null) { q.Enqueue(root.right); } h = h - 1; // Pop out the values from queue q.Dequeue(); } i = -2; // Check if the value are not // sorted then break the loop if (t == false) { break; } } if (t) Console.Write("true" + "\n"); else Console.Write("false" + "\n");}// Driver codepublic static void Main(String[] args){ node root = newnode(10); root.right = newnode(50); root.right.right = newnode(15); root.left = newnode(20); root.left.left = newnode(50); root.left.right = newnode(23); root.right.left = newnode(10); func(root);}}// This code is contributed by gauravrajput1 |
Javascript
<script> // Javascript implementation to check if the left // view of the given tree is sorted or not // Binary Tree Node class node { constructor(key) { this.left = null; this.right = null; this.val = key; } } // Utility function to create a new node function newnode(key) { let temp = new node(key); return temp; } // Function to find left view // and check if it is sorted function func(root) { // Queue to hold values let q = []; // Variable to check whether // level order is sorted or not let t = true; q.push(root); let i = -1, j = -1; // Iterate until the queue is empty while (q.length > 0) { let h = q.length; // Traverse every level in tree while (h > 0) { root = q[0]; // Variable for initial level if (i == -1) { j = root.val; } // Checking values are sorted or not if (i == -2) { if (j <= root.val) { j = root.val; i = -3; } else { t = false; break; } } // Push left value if it is not null if (root.left != null) { q.push(root.left); } // Push right value if it is not null if (root.right != null) { q.push(root.right); } h = h - 1; // Pop out the values from queue q.shift(); } i = -2; // Check if the value are not // sorted then break the loop if (t == false) { break; } } if (t) document.write("true" + "</br>"); else document.write("false" + "</br>"); } let root = newnode(10); root.right = newnode(50); root.right.right = newnode(15); root.left = newnode(20); root.left.left = newnode(50); root.left.right = newnode(23); root.right.left = newnode(10); func(root);// This code is contributed by decode2207.</script> |
Output:
true
Time complexity: O(N)
Auxiliary Space: O(N)
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