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# 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:

1. If both trees are empty then return 1.
2. 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.
3. 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 ``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 ``#include ` `// 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

 ``

Output

`Both trees have same structure`

Time Complexity: O(N)
Auxiliary Space: O(N)