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

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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; ` `} ` ` `  `// Fucntion 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 `

Output:

```true
```

Time complexity: O(N)

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : GauravRajput1

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.