Given two Binary Search Trees consisting of unique positive elements, we have to check whether the two BSTs contain the same set of elements or not.
Note: The structure of the two given BSTs can be different.
For example,
The above two BSTs contains same set of elements {5, 10, 12, 15, 20, 25}
Method 1: The most simple method will be to traverse first tree and store its element in a list or array. Now, traverse 2nd tree and simultaneously check if the current element is present in the list or not. If yes, then mark the element in the list as negative and check for further elements otherwise if no, then immediately terminate the traversal and print No. If all the elements of 2nd tree is present in the list and are marked negative then finally traverse the list to check if there are any non-negative elements left. If Yes then it means that the first tree had some extra element otherwise both trees consist the same set of elements.
Time Complexity: O( n * n ) , where n is the number of nodes in the BST.
Auxiliary Space: O( n ).
Method 2: This method is an optimization of the above approach. If we observe carefully, we will see that in the above approach, searching for elements in the list takes linear time. We can optimize this operation to be done in constant time using a hashmap instead of a list. We insert elements of both trees in different hash sets. Finally, we compare if both hash sets contain the same elements or not.
Below is the complete implementation of the above approach:
// CPP program to check if two BSTs contains // same set of elements #include<bits/stdc++.h> using namespace std;
// BST Node struct Node
{ int data;
struct Node* left;
struct Node* right;
}; // Utility function to create new Node Node* newNode( int val)
{ Node* temp = new Node;
temp->data = val;
temp->left = temp->right = NULL;
return temp;
} // function to insert elements of the // tree to map m void insertToHash(Node* root, unordered_set< int > &s)
{ if (!root)
return ;
insertToHash(root->left, s);
s.insert(root->data);
insertToHash(root->right, s);
} // function to check if the two BSTs contain // same set of elements bool checkBSTs(Node* root1, Node* root2)
{ // Base cases
if (!root1 && !root2)
return true ;
if ((root1 && !root2) || (!root1 && root2))
return false ;
// Create two hash sets and store
// elements both BSTs in them.
unordered_set< int > s1, s2;
insertToHash(root1, s1);
insertToHash(root2, s2);
// Return true if both hash sets
// contain same elements.
return (s1 == s2);
} // Driver program to check above functions int main()
{ // First BST
Node* root1 = newNode(15);
root1->left = newNode(10);
root1->right = newNode(20);
root1->left->left = newNode(5);
root1->left->right = newNode(12);
root1->right->right = newNode(25);
// Second BST
Node* root2 = newNode(15);
root2->left = newNode(12);
root2->right = newNode(20);
root2->left->left = newNode(5);
root2->left->left->right = newNode(10);
root2->right->right = newNode(25);
// check if two BSTs have same set of elements
if (checkBSTs(root1, root2))
cout << "YES" ;
else
cout << "NO" ;
return 0;
} |
// Java program to check if two BSTs // contain same set of elements import java.util.*;
class GFG {
// BST Node
static class Node {
int data;
Node left;
Node right;
};
// Utility function to create new Node
static Node newNode( int val)
{
Node temp = new Node();
temp.data = val;
temp.left = temp.right = null ;
return temp;
}
// function to insert elements
// of the tree to map m
static void storeInorder(Node root, Vector<Integer> v)
{
if (root == null )
return ;
storeInorder(root.left, v);
v.add(root.data);
storeInorder(root.right, v);
}
// function to check if the two BSTs
// contain same set of elements
static boolean checkBSTs(Node root1, Node root2)
{
// Base cases
if (root1 == null && root2 == null )
return true ;
if ((root1 == null && root2 != null )
|| (root1 != null && root2 == null ))
return false ;
// Create two vectors and store
// inorder traversals of both BSTs
// in them.
Vector<Integer> v1 = new Vector<Integer>();
Vector<Integer> v2 = new Vector<Integer>();
storeInorder(root1, v1);
storeInorder(root2, v2);
// Return true if both vectors are
// identical
return (v1.hashCode() == v2.hashCode());
}
// Driver Code
public static void main(String[] args)
{
// First BST
Node root1 = newNode( 15 );
root1.left = newNode( 10 );
root1.right = newNode( 20 );
root1.left.left = newNode( 5 );
root1.left.right = newNode( 12 );
root1.right.right = newNode( 25 );
// Second BST
Node root2 = newNode( 15 );
root2.left = newNode( 12 );
root2.right = newNode( 20 );
root2.left.left = newNode( 5 );
root2.left.left.right = newNode( 10 );
root2.right.right = newNode( 25 );
// check if two BSTs have same set of elements
if (checkBSTs(root1, root2))
System.out.print( "YES" );
else
System.out.print( "NO" );
}
} // This code is contributed by Rajput-Ji // Code corrected by Prithi_Raj |
# Python3 program to check if two BSTs contains # same set of elements # BST Node class Node:
def __init__( self ):
self .data = 0
self .left = None
self .right = None
# Utility function to create Node def Node_(val1):
temp = Node()
temp.data = val1
temp.left = temp.right = None
return temp
v = []
# function to insert elements of the # tree to map m def storeInorder(root):
if (root = = None ):
return
storeInorder(root.left)
v.append(root.data)
storeInorder(root.right)
# function to check if the two BSTs contain # same set of elements def checkBSTs(root1, root2):
# Base cases
if (root1 = = None and root2 = = None ) :
return True
if ((root1 = = None and root2 ! = None ) or \
(root1 ! = None and root2 = = None )):
return False
# Create two hash sets and store
# elements both BSTs in them.
v1 = []
v2 = []
v = v1
storeInorder(root1)
v1 = v
v = v2
storeInorder(root2)
v2 = v
# Return True if both hash sets
# contain same elements.
return (v1 = = v2)
# Driver code # First BST root1 = Node_( 15 )
root1.left = Node_( 10 )
root1.right = Node_( 20 )
root1.left.left = Node_( 5 )
root1.left.right = Node_( 12 )
root1.right.right = Node_( 25 )
# Second BST root2 = Node_( 15 )
root2.left = Node_( 12 )
root2.right = Node_( 20 )
root2.left.left = Node_( 5 )
root2.left.left.right = Node_( 10 )
root2.right.right = Node_( 25 )
# check if two BSTs have same set of elements if (checkBSTs(root1, root2)):
print ( "YES" )
else :
print ( "NO" )
# This code is contributed by SHUBHAMSINGH10 |
// C# program to check if two BSTs // contain same set of elements using System;
using System.Collections.Generic;
class GFG
{ // BST Node class Node
{ public int data;
public Node left;
public Node right;
}; // Utility function to create new Node static Node newNode( int val)
{ Node temp = new Node();
temp.data = val;
temp.left = temp.right = null ;
return temp;
} // function to insert elements // of the tree to map m static void storeInorder(Node root,
List< int > v)
{ if (root == null )
return ;
storeInorder(root.left, v);
v.Add(root.data);
storeInorder(root.right, v);
} // function to check if the two BSTs // contain same set of elements static bool checkBSTs(Node root1,
Node root2)
{ // Base cases
if (root1 == null && root2 == null )
return true ;
if ((root1 == null && root2 != null ) ||
(root1 != null && root2 == null ))
return false ;
// Create two vectors and store
// inorder traversals of both BSTs
// in them.
List< int > v1 = new List< int >();
List< int > v2 = new List< int >();
storeInorder(root1, v1);
storeInorder(root2, v2);
// Return true if both vectors are
// identical
return (v1 == v2);
} // Driver Code public static void Main(String[] args)
{ // First BST
Node root1 = newNode(15);
root1.left = newNode(10);
root1.right = newNode(20);
root1.left.left = newNode(5);
root1.left.right = newNode(12);
root1.right.right = newNode(25);
// Second BST
Node root2 = newNode(15);
root2.left = newNode(12);
root2.right = newNode(20);
root2.left.left = newNode(5);
root2.left.left.right = newNode(10);
root2.right.right = newNode(25);
// check if two BSTs have
// same set of elements
if (checkBSTs(root1, root2))
Console.Write( "YES" );
else
Console.Write( "NO" );
} } // This code is contributed by Rajput-Ji |
<script> // JavaScript program to check if two BSTs // contain same set of elements // BST Node
class Node { constructor() {
this .data = 0;
this .prev = null ;
this .next = null ;
}
} // Utility function to create new Node
function newNode(val)
{
var temp = new Node();
temp.data = val;
temp.left = temp.right = null ;
return temp;
}
// function to insert elements
// of the tree to map m
function storeInorder(root, v) {
if (root == null )
return ;
storeInorder(root.left, v);
v.push(root.data);
storeInorder(root.right, v);
}
// function to check if the two BSTs
// contain same set of elements
function checkBSTs(root1, root2)
{
// Base cases
if (root1 == null && root2 == null )
return true ;
if ((root1 == null && root2 != null ) ||
(root1 != null && root2 == null ))
return false ;
// Create two vectors and store
// inorder traversals of both BSTs
// in them.
var v1 = [];
var v2 = [];
storeInorder(root1, v1);
storeInorder(root2, v2);
// Return true if both vectors are
// identical
return (v1 == v2);
}
// Driver Code
// First BST
var root1 = newNode(15);
root1.left = newNode(10);
root1.right = newNode(20);
root1.left.left = newNode(5);
root1.left.right = newNode(12);
root1.right.right = newNode(25);
// Second BST
var root2 = newNode(15);
root2.left = newNode(12);
root2.right = newNode(20);
root2.left.left = newNode(5);
root2.left.left.right = newNode(10);
root2.right.right = newNode(25);
// check if two BSTs have same set of elements
if (checkBSTs(root1, root2))
document.write( "YES" );
else
document.write( "NO" );
// This code is contributed by Rajput-Ji </script> |
YES
Time Complexity: O( n ), where n is the number of nodes in the trees.
Auxiliary Space: O( n ).
Method 3: We know about an interesting property of BST that inorder traversal of a BST generates a sorted array. So we can do inorder traversals of both the BSTs and generate two arrays and finally, we can compare these two arrays. If both of the arrays are the same then the BSTs have the same set of elements otherwise not.
Implementation:
// CPP program to check if two BSTs contains // same set of elements #include<bits/stdc++.h> using namespace std;
// BST Node struct Node
{ int data;
struct Node* left;
struct Node* right;
}; // Utility function to create new Node Node* newNode( int val)
{ Node* temp = new Node;
temp->data = val;
temp->left = temp->right = NULL;
return temp;
} // function to insert elements of the // tree to map m void storeInorder(Node* root, vector< int > &v)
{ if (!root)
return ;
storeInorder(root->left, v);
v.push_back(root->data);
storeInorder(root->right, v);
} // function to check if the two BSTs contain // same set of elements bool checkBSTs(Node* root1, Node* root2)
{ // Base cases
if (!root1 && !root2)
return true ;
if ((root1 && !root2) || (!root1 && root2))
return false ;
// Create two vectors and store
// inorder traversals of both BSTs
// in them.
vector< int > v1, v2;
storeInorder(root1, v1);
storeInorder(root2, v2);
// Return true if both vectors are
// identical
return (v1 == v2);
} // Driver program to check above functions int main()
{ // First BST
Node* root1 = newNode(15);
root1->left = newNode(10);
root1->right = newNode(20);
root1->left->left = newNode(5);
root1->left->right = newNode(12);
root1->right->right = newNode(25);
// Second BST
Node* root2 = newNode(15);
root2->left = newNode(12);
root2->right = newNode(20);
root2->left->left = newNode(5);
root2->left->left->right = newNode(10);
root2->right->right = newNode(25);
// check if two BSTs have same set of elements
if (checkBSTs(root1, root2))
cout << "YES" ;
else
cout << "NO" ;
return 0;
} |
// Java program to check if two BSTs // contain same set of elements import java.util.*;
class GFG {
// BST Node
static class Node {
int data;
Node left;
Node right;
};
// Utility function to create new Node
static Node newNode( int val)
{
Node temp = new Node();
temp.data = val;
temp.left = temp.right = null ;
return temp;
}
// function to insert elements
// of the tree to map m
static void storeInorder(Node root, Vector<Integer> v)
{
if (root == null )
return ;
storeInorder(root.left, v);
v.add(root.data);
storeInorder(root.right, v);
}
// function to check if the two BSTs
// contain same set of elements
static boolean checkBSTs(Node root1, Node root2)
{
// Base cases
if (root1 == null && root2 == null )
return true ;
if ((root1 == null && root2 != null )
|| (root1 != null && root2 == null ))
return false ;
// Create two vectors and store
// inorder traversals of both BSTs
// in them.
Vector<Integer> v1 = new Vector<Integer>();
Vector<Integer> v2 = new Vector<Integer>();
storeInorder(root1, v1);
storeInorder(root2, v2);
// Return true if both vectors are
// identical
return (v1.hashCode() == v2.hashCode());
}
// Driver Code
public static void main(String[] args)
{
// First BST
Node root1 = newNode( 15 );
root1.left = newNode( 10 );
root1.right = newNode( 20 );
root1.left.left = newNode( 5 );
root1.left.right = newNode( 12 );
root1.right.right = newNode( 25 );
// Second BST
Node root2 = newNode( 15 );
root2.left = newNode( 12 );
root2.right = newNode( 20 );
root2.left.left = newNode( 5 );
root2.left.left.right = newNode( 10 );
root2.right.right = newNode( 25 );
// check if two BSTs have same set of elements
if (checkBSTs(root1, root2))
System.out.print( "YES" );
else
System.out.print( "NO" );
}
} // This code is contributed by Rajput-Ji // Code corrected by Prithi_Raj |
# Python3 program to check if two BSTs contains # same set of elements # BST Node class Node:
def __init__( self ):
self .data = 0
self .left = None
self .right = None
# Utility function to create Node def Node_(val1):
temp = Node()
temp.data = val1
temp.left = temp.right = None
return temp
v = []
# function to insert elements of the # tree to map m def storeInorder(root):
if (root = = None ):
return
storeInorder(root.left)
v.append(root.data)
storeInorder(root.right)
# function to check if the two BSTs contain # same set of elements def checkBSTs(root1, root2):
# Base cases
if (root1 = = None and root2 = = None ) :
return True
if ((root1 = = None and root2 ! = None ) or \
(root1 ! = None and root2 = = None )):
return False
# Create two hash sets and store
# elements both BSTs in them.
v1 = []
v2 = []
v = v1
storeInorder(root1)
v1 = v
v = v2
storeInorder(root2)
v2 = v
# Return True if both hash sets
# contain same elements.
return (v1 = = v2)
# Driver code # First BST root1 = Node_( 15 )
root1.left = Node_( 10 )
root1.right = Node_( 20 )
root1.left.left = Node_( 5 )
root1.left.right = Node_( 12 )
root1.right.right = Node_( 25 )
# Second BST root2 = Node_( 15 )
root2.left = Node_( 12 )
root2.right = Node_( 20 )
root2.left.left = Node_( 5 )
root2.left.left.right = Node_( 10 )
root2.right.right = Node_( 25 )
# check if two BSTs have same set of elements if (checkBSTs(root1, root2)):
print ( "YES" )
else :
print ( "NO" )
# This code is contributed by SHUBHAMSINGH10 |
// C# program to check if two BSTs // contain same set of elements using System;
using System.Collections.Generic;
class GFG
{ // BST Node class Node
{ public int data;
public Node left;
public Node right;
}; // Utility function to create new Node static Node newNode( int val)
{ Node temp = new Node();
temp.data = val;
temp.left = temp.right = null ;
return temp;
} // function to insert elements // of the tree to map m static void storeInorder(Node root,
List< int > v)
{ if (root == null )
return ;
storeInorder(root.left, v);
v.Add(root.data);
storeInorder(root.right, v);
} // function to check if the two BSTs // contain same set of elements static bool checkBSTs(Node root1,
Node root2)
{ // Base cases
if (root1 == null && root2 == null )
return true ;
if ((root1 == null && root2 != null ) ||
(root1 != null && root2 == null ))
return false ;
// Create two vectors and store
// inorder traversals of both BSTs
// in them.
List< int > v1 = new List< int >();
List< int > v2 = new List< int >();
storeInorder(root1, v1);
storeInorder(root2, v2);
// Return true if both vectors are
// identical
return (v1 == v2);
} // Driver Code public static void Main(String[] args)
{ // First BST
Node root1 = newNode(15);
root1.left = newNode(10);
root1.right = newNode(20);
root1.left.left = newNode(5);
root1.left.right = newNode(12);
root1.right.right = newNode(25);
// Second BST
Node root2 = newNode(15);
root2.left = newNode(12);
root2.right = newNode(20);
root2.left.left = newNode(5);
root2.left.left.right = newNode(10);
root2.right.right = newNode(25);
// check if two BSTs have
// same set of elements
if (checkBSTs(root1, root2))
Console.Write( "YES" );
else
Console.Write( "NO" );
} } // This code is contributed by Rajput-Ji |
<script> // JavaScript program to check if two BSTs // contain same set of elements // BST Node
class Node { constructor() {
this .data = 0;
this .prev = null ;
this .next = null ;
}
} // Utility function to create new Node
function newNode(val)
{
var temp = new Node();
temp.data = val;
temp.left = temp.right = null ;
return temp;
}
// function to insert elements
// of the tree to map m
function storeInorder(root, v) {
if (root == null )
return ;
storeInorder(root.left, v);
v.push(root.data);
storeInorder(root.right, v);
}
// function to check if the two BSTs
// contain same set of elements
function checkBSTs(root1, root2)
{
// Base cases
if (root1 == null && root2 == null )
return true ;
if ((root1 == null && root2 != null ) ||
(root1 != null && root2 == null ))
return false ;
// Create two vectors and store
// inorder traversals of both BSTs
// in them.
var v1 = [];
var v2 = [];
storeInorder(root1, v1);
storeInorder(root2, v2);
// Return true if both vectors are
// identical
return (v1 == v2);
}
// Driver Code
// First BST
var root1 = newNode(15);
root1.left = newNode(10);
root1.right = newNode(20);
root1.left.left = newNode(5);
root1.left.right = newNode(12);
root1.right.right = newNode(25);
// Second BST
var root2 = newNode(15);
root2.left = newNode(12);
root2.right = newNode(20);
root2.left.left = newNode(5);
root2.left.left.right = newNode(10);
root2.right.right = newNode(25);
// check if two BSTs have same set of elements
if (checkBSTs(root1, root2))
document.write( "YES" );
else
document.write( "NO" );
// This code is contributed by Rajput-Ji </script> |
YES
Time Complexity: O( n ).
Auxiliary Space: O( n ).