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++ 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 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 |
# Python implementation for the above approach # Binary Tree Node class Node:
def __init__( self , key):
self .left = None
self .right = None
self .val = key
# Utility function to create a new node def new_node(key):
temp = Node(key)
return temp
# Function to find left view # and check if it is sorted def 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 code root = 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# 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 |
<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> |
true
Time complexity: O(N)
Auxiliary Space: O(N)