Check if two trees have same structure
Given two binary trees. The task is to write a program to check if the two trees are identical in structure.
In the above figure both of the trees, Tree1 and Tree2 are identical in structure. That is, they have the same structure.
Note: This problem is different from Check if two trees are identical as here we need to compare only the structures of the two trees and not the values at their nodes.
The idea is to traverse both trees simultaneously following the same paths and keep checking if a node exists for both the trees or not.
Algorithm:
- If both trees are empty then return 1.
- Else If both trees are non-empty:
- Check left subtrees recursively i.e., call isSameStructure(tree1->left_subtree, tree2->left_subtree)
- Check right subtrees recursively i.e., call isSameStructure(tree1->right_subtree, tree2->right_subtree)
- If the value returned in above two steps are true then return 1.
- Else return 0 (one is empty and other is not).
Below is the implementation of above algorithm:
C++
// C++ program to check if two trees have // same structure #include <iostream> using namespace std; // A binary tree node has data, pointer to left child // and a pointer to right child struct Node { int data; struct Node* left; struct Node* right; }; // Helper function that allocates a new node with the // given data and NULL left and right pointers. Node* newNode( int data) { Node* node = new Node; node->data = data; node->left = NULL; node->right = NULL; return (node); } // Function to check if two trees have same // structure int isSameStructure(Node* a, Node* b) { // 1. both empty if (a==NULL && b==NULL) return 1; // 2. both non-empty -> compare them if (a!=NULL && b!=NULL) { return ( isSameStructure(a->left, b->left) && isSameStructure(a->right, b->right) ); } // 3. one empty, one not -> false return 0; } // Driver code int main() { Node *root1 = newNode(10); Node *root2 = newNode(100); root1->left = newNode(7); root1->right = newNode(15); root1->left->left = newNode(4); root1->left->right = newNode(9); root1->right->right = newNode(20); root2->left = newNode(70); root2->right = newNode(150); root2->left->left = newNode(40); root2->left->right = newNode(90); root2->right->right = newNode(200); if (isSameStructure(root1, root2)) printf ( "Both trees have same structure" ); else printf ( "Trees do not have same structure" ); return 0; } // This code is contributed by aditya kumar (adityakumar129) |
C
// C++ program to check if two trees have // same structure #include <stdio.h> #include <stdlib.h> // A binary tree node has data, pointer to left child // and a pointer to right child typedef struct Node { int data; struct Node* left; struct Node* right; } Node; // Helper function that allocates a new node with the // given data and NULL left and right pointers. Node* newNode( int data) { Node* node = (Node*) malloc ( sizeof (Node)); node->data = data; node->left = NULL; node->right = NULL; return (node); } // Function to check if two trees have same // structure int isSameStructure(Node* a, Node* b) { // 1. both empty if (a == NULL && b == NULL) return 1; // 2. both non-empty -> compare them if (a != NULL && b != NULL) { return (isSameStructure(a->left, b->left) && isSameStructure(a->right, b->right)); } // 3. one empty, one not -> false return 0; } // Driver code int main() { Node* root1 = newNode(10); Node* root2 = newNode(100); root1->left = newNode(7); root1->right = newNode(15); root1->left->left = newNode(4); root1->left->right = newNode(9); root1->right->right = newNode(20); root2->left = newNode(70); root2->right = newNode(150); root2->left->left = newNode(40); root2->left->right = newNode(90); root2->right->right = newNode(200); if (isSameStructure(root1, root2)) printf ( "Both trees have same structure" ); else printf ( "Trees do not have same structure" ); return 0; } // This code is contributed by aditya kumar (adityakumar129) |
Java
// Java program to check if two trees have // same structure class GFG { // A binary tree node has data, // pointer to left child and // a pointer to right child static class Node { int data; Node left; Node right; }; // Helper function that allocates a new node // with the given data and null left // and right pointers. static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = null ; node.right = null ; return (node); } // Function to check if two trees // have same structure static boolean isSameStructure(Node a, Node b) { // 1. both empty if (a == null && b == null ) return true ; // 2. both non-empty . compare them if (a != null && b != null ) { return ( isSameStructure(a.left, b.left) && isSameStructure(a.right, b.right) ); } // 3. one empty, one not . false return false ; } // Driver code public static void main(String args[]) { Node root1 = newNode( 10 ); Node root2 = newNode( 100 ); root1.left = newNode( 7 ); root1.right = newNode( 15 ); root1.left.left = newNode( 4 ); root1.left.right = newNode( 9 ); root1.right.right = newNode( 20 ); root2.left = newNode( 70 ); root2.right = newNode( 150 ); root2.left.left = newNode( 40 ); root2.left.right = newNode( 90 ); root2.right.right = newNode( 200 ); if (isSameStructure(root1, root2)) System.out.printf( "Both trees have same structure" ); else System.out.printf( "Trees do not have same structure" ); } } // This code is contributed by Arnab Kundu |
Python3
# Python3 program to check if two trees have # same structure # A binary tree node has data, pointer to left child # and a pointer to right child class Node: def __init__( self , data): self .left = None self .right = None self .data = data # Helper function that allocates a new node with the # given data and None left and right pointers. def newNode(data): node = Node(data) return node # Function to check if two trees have same # structure def isSameStructure(a, b): # 1. both empty if (a = = None and b = = None ): return 1 ; # 2. both non-empty . compare them if (a ! = None and b ! = None ): return ( isSameStructure(a.left, b.left) and isSameStructure(a.right, b.right)) # 3. one empty, one not . false return 0 ; # Driver code if __name__ = = '__main__' : root1 = newNode( 10 ); root2 = newNode( 100 ); root1.left = newNode( 7 ); root1.right = newNode( 15 ); root1.left.left = newNode( 4 ); root1.left.right = newNode( 9 ); root1.right.right = newNode( 20 ); root2.left = newNode( 70 ); root2.right = newNode( 150 ); root2.left.left = newNode( 40 ); root2.left.right = newNode( 90 ); root2.right.right = newNode( 200 ); if (isSameStructure(root1, root2)): print ( "Both trees have same structure" ); else : print ( "Trees do not have same structure" ); # This code is contributed by rutvik_56 |
C#
// C# program to check if two trees // have same structure using System; class GFG { // A binary tree node has data, // pointer to left child and // a pointer to right child public class Node { public int data; public Node left; public Node right; }; // Helper function that allocates a new node // with the given data and null left // and right pointers. static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = null ; node.right = null ; return (node); } // Function to check if two trees // have same structure static Boolean isSameStructure(Node a, Node b) { // 1. both empty if (a == null && b == null ) return true ; // 2. both non-empty . compare them if (a != null && b != null ) { return ( isSameStructure(a.left, b.left) && isSameStructure(a.right, b.right) ); } // 3. one empty, one not . false return false ; } // Driver code public static void Main(String []args) { Node root1 = newNode(10); Node root2 = newNode(100); root1.left = newNode(7); root1.right = newNode(15); root1.left.left = newNode(4); root1.left.right = newNode(9); root1.right.right = newNode(20); root2.left = newNode(70); root2.right = newNode(150); root2.left.left = newNode(40); root2.left.right = newNode(90); root2.right.right = newNode(200); if (isSameStructure(root1, root2)) Console.Write( "Both trees have " + "same structure" ); else Console.Write( "Trees do not have" + " same structure" ); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript program to check if two trees // have same structure // A binary tree node has data, // pointer to left child and // a pointer to right child class Node { constructor() { this .data = 0; this .left = null ; this .right = null ; } }; // Helper function that allocates a new node // with the given data and null left // and right pointers. function newNode(data) { var node = new Node(); node.data = data; node.left = null ; node.right = null ; return node; } // Function to check if two trees // have same structure function isSameStructure(a, b) { // 1. both empty if (a == null && b == null ) return true ; // 2. both non-empty . compare them if (a != null && b != null ) { return isSameStructure(a.left, b.left) && isSameStructure(a.right, b.right) ; } // 3. one empty, one not . false return false ; } // Driver code var root1 = newNode(10); var root2 = newNode(100); root1.left = newNode(7); root1.right = newNode(15); root1.left.left = newNode(4); root1.left.right = newNode(9); root1.right.right = newNode(20); root2.left = newNode(70); root2.right = newNode(150); root2.left.left = newNode(40); root2.left.right = newNode(90); root2.right.right = newNode(200); if (isSameStructure(root1, root2)) document.write( "Both trees have " + "same structure" ); else document.write( "Trees do not have" + " same structure" ); </script> |
Output
Both trees have same structure
Time Complexity: O(N)
Auxiliary Space: O(N)
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