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 <bits/stdc++.h>
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<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 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<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 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<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 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

<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)

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