# Check if the Left View of the given tree is sorted or not

• Last Updated : 23 Jun, 2021

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 using namespace std;   // Binary Tree Node struct node {     int val;     struct node *right, *left; };   // Utility function to create a new node struct 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 sorted void func(node* root) {     // queue to hold values     queue 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 code int 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 not import java.util.*;   class GFG{   // Binary Tree Node static class node {     int val;     node right, left; };   // Utility function to create a new node static 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 sorted static void func(node root) {           // Queue to hold values     Queue 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 code public 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

## C#

 // C# implementation to check if the left // view of the given tree is sorted or not using System; using System.Collections.Generic;   class GFG{   // Binary Tree Node class node {     public int val;     public node right, left; };   // Utility function to create a new node static 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 sorted static void func(node root) {           // Queue to hold values     Queue q = new Queue();       // 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 code public 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



Output:

true

Time complexity: O(N)

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