Check whether the two Binary Search Trees are Identical or Not
Given the root nodes of the two binary search trees. The task is to print “Both BSTs are identical” if the two Binary Search Trees are identical else print “Both BSTs are identical”. Two trees are identical if they are identical structurally and nodes have the same values.
Tree1: 5 / \ 3 8 / \ 2 4 Tree2: 5 / \ 3 8 / \ 2 4
Here in both the trees the structure is same and the nodes having same value so we can say that Both BSTs are identical.
To identify if two trees are identical, we need to traverse both trees simultaneously, and while traversing we need to compare data and children of the trees.
Below is the step by step algorithm to check if two BSTs are identical:
- If both trees are empty then return 1.
- Else If both trees are non -empty
- Check data of the root nodes (tree1->data == tree2->data)
- Check left subtrees recursively i.e., call sameTree(tree1->left_subtree, tree2->left_subtree)
- Check right subtrees recursively i.e., call sameTree(tree1->right_subtree, tree2->right_subtree)
- If the values returned in the above three steps are true then return 1.
- Else return 0 (one is empty and other is not).
Below is the implementation of the above approach:
C++
// C++ program to check if two BSTs // are identical #include <iostream> using namespace std; // BST node struct Node { int data; struct Node* left; struct Node* right; }; // Utility function to create a new Node struct Node* newNode( int data) { struct Node* node = ( struct Node*) malloc ( sizeof ( struct Node)); node->data = data; node->left = NULL; node->right = NULL; return node; } // Function to perform inorder traversal void inorder(Node* root) { if (root == NULL) return ; inorder(root->left); cout << root->data << " " ; inorder(root->right); } // Function to check if two BSTs // are identical int isIdentical(Node* root1, Node* root2) { // Check if both the trees are empty if (root1 == NULL && root2 == NULL) return 1; // If any one of the tree is non-empty // and other is empty, return false else if (root1 == NULL || root2 == NULL) return 0; else { // Check if current data of both trees equal // and recursively check for left and right subtrees if (root1->data == root2->data && isIdentical(root1->left, root2->left) && isIdentical(root1->right, root2->right)) return 1; else return 0; } } // Driver code int main() { struct Node* root1 = newNode(5); struct Node* root2 = newNode(5); root1->left = newNode(3); root1->right = newNode(8); root1->left->left = newNode(2); root1->left->right = newNode(4); root2->left = newNode(3); root2->right = newNode(8); root2->left->left = newNode(2); root2->left->right = newNode(4); if (isIdentical(root1, root2)) cout << "Both BSTs are identical" ; else cout << "BSTs are not identical" ; return 0; } |
C
// C program to check if two BSTs // are identical #include <stdio.h> #include <stdlib.h> // BST node struct Node { int data; struct Node* left; struct Node* right; }; // Utility function to create a new Node struct Node* newNode( int data) { struct Node* node = ( struct Node*) malloc ( sizeof ( struct Node)); node->data = data; node->left = NULL; node->right = NULL; return node; } // Function to perform inorder traversal void inorder( struct Node* root) { if (root == NULL) return ; inorder(root->left); printf ( "%d " ,root->data); inorder(root->right); } // Function to check if two BSTs // are identical int isIdentical( struct Node* root1, struct Node* root2) { // Check if both the trees are empty if (root1 == NULL && root2 == NULL) return 1; // If any one of the tree is non-empty // and other is empty, return false else if (root1 == NULL || root2 == NULL) return 0; else { // Check if current data of both trees equal // and recursively check for left and right subtrees if (root1->data == root2->data && isIdentical(root1->left, root2->left) && isIdentical(root1->right, root2->right)) return 1; else return 0; } } // Driver code int main() { struct Node* root1 = newNode(5); struct Node* root2 = newNode(5); root1->left = newNode(3); root1->right = newNode(8); root1->left->left = newNode(2); root1->left->right = newNode(4); root2->left = newNode(3); root2->right = newNode(8); root2->left->left = newNode(2); root2->left->right = newNode(4); if (isIdentical(root1, root2)) printf ( "Both BSTs are identical" ); else printf ( "BSTs are not identical" ) ; return 0; } // This code is contributed by sanskar84. |
Java
// Java program to check if two BSTs // are identical class GFG { // BST node static class Node { int data; Node left; Node right; }; // Utility function to create a new Node static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = null ; node.right = null ; return node; } // Function to perform inorder traversal static void inorder(Node root) { if (root == null ) return ; inorder(root.left); System.out.print(root.data + " " ); inorder(root.right); } // Function to check if two BSTs // are identical static int isIdentical(Node root1, Node root2) { // Check if both the trees are empty if (root1 == null && root2 == null ) return 1 ; // If any one of the tree is non-empty // and other is empty, return false else if (root1 != null && root2 == null ) return 0 ; else if (root1 == null && root2 != null ) return 0 ; else { // Check if current data of both trees equal // and recursively check for left and right subtrees if (root1.data == root2.data && isIdentical(root1.left, root2.left) == 1 && isIdentical(root1.right, root2.right) == 1 ) return 1 ; else return 0 ; } } // Driver code public static void main(String[] args) { Node root1 = newNode( 5 ); Node root2 = newNode( 5 ); root1.left = newNode( 3 ); root1.right = newNode( 8 ); root1.left.left = newNode( 2 ); root1.left.right = newNode( 4 ); root2.left = newNode( 3 ); root2.right = newNode( 8 ); root2.left.left = newNode( 2 ); root2.left.right = newNode( 4 ); if (isIdentical(root1, root2)== 1 ) System.out.print( "Both BSTs are identical" ); else System.out.print( "BSTs are not identical" ); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to construct all unique # BSTs for keys from 1 to n # Binary Tree Node """ A utility function to create a new BST node """ class newNode: # Construct to create a newNode def __init__( self , data): self .data = data self .left = None self .right = None # Function to perform inorder traversal def inorder(root) : if (root = = None ): return inorder(root.left) print (root.data, end = " " ) inorder(root.right) # Function to check if two BSTs # are identical def isIdentical(root1, root2) : # Check if both the trees are empty if (root1 = = None and root2 = = None ) : return 1 # If any one of the tree is non-empty # and other is empty, return false elif (root1 ! = None and root2 = = None ) : return 0 elif (root1 = = None and root2 ! = None ) : return 0 else : # Check if current data of both trees # equal and recursively check for left # and right subtrees if (root1.data = = root2.data and isIdentical(root1.left, root2.left) and isIdentical(root1.right, root2.right)) : return 1 else : return 0 # Driver Code if __name__ = = '__main__' : root1 = newNode( 5 ) root2 = newNode( 5 ) root1.left = newNode( 3 ) root1.right = newNode( 8 ) root1.left.left = newNode( 2 ) root1.left.right = newNode( 4 ) root2.left = newNode( 3 ) root2.right = newNode( 8 ) root2.left.left = newNode( 2 ) root2.left.right = newNode( 4 ) if (isIdentical(root1, root2)): print ( "Both BSTs are identical" ) else : print ( "BSTs are not identical" ) # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to check if two BSTs // are identical using System; class GFG { // BST node class Node { public int data; public Node left; public Node right; }; // Utility function to create a new Node static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = null ; node.right = null ; return node; } // Function to perform inorder traversal static void inorder(Node root) { if (root == null ) return ; inorder(root.left); Console.Write(root.data + " " ); inorder(root.right); } // Function to check if two BSTs // are identical static int isIdentical(Node root1, Node root2) { // Check if both the trees are empty if (root1 == null && root2 == null ) return 1; // If any one of the tree is non-empty // and other is empty, return false else if (root1 != null && root2 == null ) return 0; else if (root1 == null && root2 != null ) return 0; else { // Check if current data of both trees equal // and recursively check for left and right subtrees if (root1.data == root2.data && isIdentical(root1.left, root2.left) == 1 && isIdentical(root1.right, root2.right) == 1) return 1; else return 0; } } // Driver code public static void Main(String[] args) { Node root1 = newNode(5); Node root2 = newNode(5); root1.left = newNode(3); root1.right = newNode(8); root1.left.left = newNode(2); root1.left.right = newNode(4); root2.left = newNode(3); root2.right = newNode(8); root2.left.left = newNode(2); root2.left.right = newNode(4); if (isIdentical(root1, root2) == 1) Console.Write( "Both BSTs are identical" ); else Console.Write( "BSTs are not identical" ); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript program to check if two BSTs // are identical // BST node class Node { // Utility function to create a new Node constructor(data) { this .data = data; this .left = this .right = null ; } } // Function to perform inorder traversal function inorder(root) { if (root == null ) return ; inorder(root.left); document.write(root.data + " " ); inorder(root.right); } // Function to check if two BSTs // are identical function isIdentical(root1,root2) { // Check if both the trees are empty if (root1 == null && root2 == null ) return 1; // If any one of the tree is non-empty // and other is empty, return false else if (root1 != null && root2 == null ) return 0; else if (root1 == null && root2 != null ) return 0; else { // Check if current data of both trees equal // and recursively check for left and right subtrees if (root1.data == root2.data && isIdentical(root1.left, root2.left) == 1 && isIdentical(root1.right, root2.right) == 1) return 1; else return 0; } } // Driver code let root1 = new Node(5); let root2 = new Node(5); root1.left = new Node(3); root1.right = new Node(8); root1.left.left = new Node(2); root1.left.right = new Node(4); root2.left = new Node(3); root2.right = new Node(8); root2.left.left = new Node(2); root2.left.right = new Node(4); if (isIdentical(root1, root2)==1) document.write( "Both BSTs are identical" ); else document.write( "BSTs are not identical" ); // This code is contributed by avanitrachhadiya2155 </script> |
Both BSTs are identical
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N)
Another Approach:
To solve the problem follow the below idea:
If two binary search trees are identical, their Preorder, Inorder and Postorder traversals will also be the same.
Below is the implementation of the above approach:
C++
// C++ program to check if two binary search trees are // identical or not #include <bits/stdc++.h> using namespace std; // A binary tree node class Node { public : int data; Node* left; Node* right; Node( int d) { this ->data = d; this ->left = NULL; this ->right = NULL; } }; // function to check inorder traversal void inOrder(Node* root, vector< int >& sol) { if (root == NULL) return ; inOrder(root->left, sol); sol.push_back(root->data); inOrder(root->right, sol); } // function to check preorder traversal void preOrder(Node* root, vector< int >& sol) { if (root == NULL) return ; sol.push_back(root->data); preOrder(root->left, sol); preOrder(root->right, sol); } // function to check postorder traversal void postOrder(Node* root, vector< int >& sol) { if (root == NULL) return ; postOrder(root->left, sol); postOrder(root->right, sol); sol.push_back(root->data); } // Function to check if two trees are identical bool isIdentical(Node* root1, Node* root2) { // Create two vector to store traversals vector< int > res1; vector< int > res2; // check inOrder inOrder(root1, res1); inOrder(root2, res2); if (res1 != res2) return false ; // clear previous result to reuse vector res1.clear(); res2.clear(); // check PreOrder preOrder(root1, res1); preOrder(root2, res2); if (res1 != res2) return false ; // clear previous result to reuse vector res1.clear(); res2.clear(); // check PostOrder postOrder(root1, res1); postOrder(root2, res2); if (res1 != res2) return false ; return true ; } // Driver code int main() { struct Node* root1 = new Node(5); struct Node* root2 = new Node(5); root1->left = new Node(3); root1->right = new Node(8); root1->left->left = new Node(2); root1->left->right = new Node(4); root2->left = new Node(3); root2->right = new Node(8); root2->left->left = new Node(2); root2->left->right = new Node(4); // Function call if (isIdentical(root1, root2)) { cout << "Both BSTs are identical." << endl; } else { cout << "BSTs are not identical." << endl; } return 0; } // This code is contributed by Yash Agarwal(yashagarwal2852002) |
Java
// Java program to check if two binary search trees are // identical or not import java.util.*; // A binary tree node class Node { int data; Node left, right; Node( int d) { this .data = d; this .left = null ; this .right = null ; } } // Main class public class Main { // function to check inorder traversal static void inOrder(Node root, ArrayList<Integer> sol) { if (root == null ) return ; inOrder(root.left, sol); sol.add(root.data); inOrder(root.right, sol); } // function to check preorder traversal static void preOrder(Node root, ArrayList<Integer> sol) { if (root == null ) return ; sol.add(root.data); preOrder(root.left, sol); preOrder(root.right, sol); } // function to check postorder traversal static void postOrder(Node root, ArrayList<Integer> sol) { if (root == null ) return ; postOrder(root.left, sol); postOrder(root.right, sol); sol.add(root.data); } // Function to check if two trees are identical static boolean isIdentical(Node root1, Node root2) { // Create two array lists to store traversals ArrayList<Integer> res1 = new ArrayList<Integer>(); ArrayList<Integer> res2 = new ArrayList<Integer>(); // check inOrder inOrder(root1, res1); inOrder(root2, res2); if (!res1.equals(res2)) return false ; // clear previous result to reuse array list res1.clear(); res2.clear(); // check PreOrder preOrder(root1, res1); preOrder(root2, res2); if (!res1.equals(res2)) return false ; // clear previous result to reuse array list res1.clear(); res2.clear(); // check PostOrder postOrder(root1, res1); postOrder(root2, res2); if (!res1.equals(res2)) return false ; return true ; } // Driver code public static void main(String args[]) { Node root1 = new Node( 5 ); Node root2 = new Node( 5 ); root1.left = new Node( 3 ); root1.right = new Node( 8 ); root1.left.left = new Node( 2 ); root1.left.right = new Node( 4 ); root2.left = new Node( 3 ); root2.right = new Node( 8 ); root2.left.left = new Node( 2 ); root2.left.right = new Node( 4 ); // Function call if (isIdentical(root1, root2)) { System.out.println( "Both BSTs are identical." ); } else { System.out.println( "BSTs are not identical." ); } } } |
Python3
# Python3 program to check if two binary search trees are identical or not # A binary tree node class Node: def __init__( self , d): self .data = d self .left = None self .right = None # function to check inorder traversal def inOrder(root, sol): if root is None : return inOrder(root.left, sol) sol.append(root.data) inOrder(root.right, sol) # function to check preorder traversal def preOrder(root, sol): if root is None : return sol.append(root.data) preOrder(root.left, sol) preOrder(root.right, sol) # function to check postorder traversal def postOrder(root, sol): if root is None : return postOrder(root.left, sol) postOrder(root.right, sol) sol.append(root.data) # Function to check if two trees are identical def isIdentical(root1, root2): # Create two list to store traversals res1 = [] res2 = [] # check inOrder inOrder(root1, res1) inOrder(root2, res2) if res1 ! = res2: return False # clear previous result to reuse list res1.clear() res2.clear() # check PreOrder preOrder(root1, res1) preOrder(root2, res2) if res1 ! = res2: return False # clear previous result to reuse list res1.clear() res2.clear() # check PostOrder postOrder(root1, res1) postOrder(root2, res2) if res1 ! = res2: return False return True # Driver code if __name__ = = "__main__" : root1 = Node( 5 ) root2 = Node( 5 ) root1.left = Node( 3 ) root1.right = Node( 8 ) root1.left.left = Node( 2 ) root1.left.right = Node( 4 ) root2.left = Node( 3 ) root2.right = Node( 8 ) root2.left.left = Node( 2 ) root2.left.right = Node( 4 ) # Function call if isIdentical(root1, root2): print ( "Both BSTs are identical." ) else : print ( "BSTs are not identical." ) # This code is contributed by divyansh2212 |
Javascript
// JavaScript Program to check if two binary search trees are // identical or not // a binary tree node class Node{ constructor(d){ this .data = d; this .left = null ; this .right = null ; } } // a function to serialize the arrays into strings and compare them const equalCheck = (a,b)=>{ return JSON.stringify(a) === JSON.stringify(b); } let res = []; // function to check inorder traversal function inOrder(root){ if (root == null ) return res; inOrder(root.left); res.push(root.data); inOrder(root.right); return res; } // function to check preorder traversal function preOrder(root){ if (root == null ) return res; res.push(root.data); preOrder(root.left); preOrder(root.right); } // function to check postorder traversal function postOrder(root){ if (root == null ) return res; postOrder(root.left); postOrder(root.right); res.push(root.data); } // function to check if two trees are idnetical function isIdentical(root1, root2){ // create two vector to store traversal let res1 = []; let res2 = []; // check inorder res1 = inOrder(root1); res = []; res2 = inOrder(root2); res = []; if (equalCheck(res1, res2) == false ) return false ; // check preorder res1 = preOrder(root1); res = []; res2 = preOrder(root2); res = []; if (equalCheck(res1, res2) == false ) return false ; // check postorder res1 = postOrder(root1); res = []; res2 = postOrder(root2); if (equalCheck(res1, res2) == false ) return false ; return true ; } // driver code let root1 = new Node(5); let root2 = new Node(5); root1.left = new Node(3); root1.right = new Node(8); root1.left.left = new Node(2); root1.left.right = new Node(4); root2.left = new Node(3); root2.right = new Node(8); root2.left.left = new Node(2); root2.left.right = new Node(4); // function call if (isIdentical(root1, root2)){ console.log( "Both BSTs are identical." ); } else { console.log( "BSTs are not identical." ); } |
C#
// C# program for the above approach using System; using System.Collections.Generic; // Binary tree node class Node { public int data; public Node left, right; public Node( int d) { this .data = d; this .left = null ; this .right = null ; } } public class MainClass { // function to check inorder traversal static void inOrder(Node root, List< int > sol) { if (root == null ) return ; inOrder(root.left, sol); sol.Add(root.data); inOrder(root.right, sol); } // function to check preorder traversal static void preOrder(Node root, List< int > sol) { if (root == null ) return ; sol.Add(root.data); preOrder(root.left, sol); preOrder(root.right, sol); } // function to check postorder traversal static void postOrder(Node root, List< int > sol) { if (root == null ) return ; postOrder(root.left, sol); postOrder(root.right, sol); sol.Add(root.data); } // Function to check if two trees are identical static bool isIdentical(Node root1, Node root2) { // Create two lists to store traversals List< int > res1 = new List< int >(); List< int > res2 = new List< int >(); // check inOrder inOrder(root1, res1); inOrder(root2, res2); if (!res1.Equals(res2)) return false ; // clear previous result to reuse list res1.Clear(); res2.Clear(); // check PreOrder preOrder(root1, res1); preOrder(root2, res2); if (!res1.Equals(res2)) return false ; // clear previous result to reuse list res1.Clear(); res2.Clear(); // check PostOrder postOrder(root1, res1); postOrder(root2, res2); if (!res1.Equals(res2)) return false ; return true ; } // Driver code static void Main() { Node root1 = new Node(5); Node root2 = new Node(5); root1.left = new Node(3); root1.right = new Node(8); root1.left.left = new Node(2); root1.left.right = new Node(4); root2.left = new Node(3); root2.right = new Node(8); root2.left.left = new Node(2); root2.left.right = new Node(4); // Function call if (!isIdentical(root1, root2)) { Console.WriteLine( "Both BSTs are identical." ); } else { Console.WriteLine( "BSTs are not identical." ); } } } // This code is contributed by codebraxnzt |
Both BSTs are identical.
Time Complexity: O(NlogN)
Auxiliary Space: O(N)