Write a function to detect if two trees are isomorphic. Two trees are called isomorphic if one of them can be obtained from other by a series of flips, i.e. by swapping left and right children of a number of nodes. Any number of nodes at any level can have their children swapped. Two empty trees are isomorphic.

For example, following two trees are isomorphic with following sub-trees flipped: 2 and 3, NULL and 6, 7 and 8.

We simultaneously traverse both trees. Let the current internal nodes of two trees being traversed be **n1 **and **n2** respectively. There are following two conditions for subtrees rooted with n1 and n2 to be isomorphic.

**1)** Data of n1 and n2 is same.

**2) **One of the following two is true for children of n1 and n2

……**a)** Left child of n1 is isomorphic to left child of n2 and right child of n1 is isomorphic to right child of n2.

……**b)** Left child of n1 is isomorphic to right child of n2 and right child of n1 is isomorphic to left child of n2.

## C++

`// A C++ program to check if two given trees are isomorphic ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `/* A binary tree node has data, pointer to left and right children */` `struct` `node ` `{ ` ` ` `int` `data; ` ` ` `struct` `node* left; ` ` ` `struct` `node* right; ` `}; ` ` ` `/* Given a binary tree, print its nodes in reverse level order */` `bool` `isIsomorphic(node* n1, node *n2) ` `{ ` ` ` `// Both roots are NULL, trees isomorphic by definition ` ` ` `if` `(n1 == NULL && n2 == NULL) ` ` ` `return` `true` `; ` ` ` ` ` `// Exactly one of the n1 and n2 is NULL, trees not isomorphic ` ` ` `if` `(n1 == NULL || n2 == NULL) ` ` ` `return` `false` `; ` ` ` ` ` `if` `(n1->data != n2->data) ` ` ` `return` `false` `; ` ` ` ` ` `// There are two possible cases for n1 and n2 to be isomorphic ` ` ` `// Case 1: The subtrees rooted at these nodes have NOT been "Flipped". ` ` ` `// Both of these subtrees have to be isomorphic, hence the && ` ` ` `// Case 2: The subtrees rooted at these nodes have been "Flipped" ` ` ` `return` ` ` `(isIsomorphic(n1->left,n2->left) && isIsomorphic(n1->right,n2->right))|| ` ` ` `(isIsomorphic(n1->left,n2->right) && isIsomorphic(n1->right,n2->left)); ` `} ` ` ` `/* Helper function that allocates a new node with the ` ` ` `given data and NULL left and right pointers. */` `node* newNode(` `int` `data) ` `{ ` ` ` `node* temp = ` `new` `node; ` ` ` `temp->data = data; ` ` ` `temp->left = NULL; ` ` ` `temp->right = NULL; ` ` ` ` ` `return` `(temp); ` `} ` ` ` `/* Driver program to test above functions*/` `int` `main() ` `{ ` ` ` `// Let us create trees shown in above diagram ` ` ` `struct` `node *n1 = newNode(1); ` ` ` `n1->left = newNode(2); ` ` ` `n1->right = newNode(3); ` ` ` `n1->left->left = newNode(4); ` ` ` `n1->left->right = newNode(5); ` ` ` `n1->right->left = newNode(6); ` ` ` `n1->left->right->left = newNode(7); ` ` ` `n1->left->right->right = newNode(8); ` ` ` ` ` `struct` `node *n2 = newNode(1); ` ` ` `n2->left = newNode(3); ` ` ` `n2->right = newNode(2); ` ` ` `n2->right->left = newNode(4); ` ` ` `n2->right->right = newNode(5); ` ` ` `n2->left->right = newNode(6); ` ` ` `n2->right->right->left = newNode(8); ` ` ` `n2->right->right->right = newNode(7); ` ` ` ` ` `if` `(isIsomorphic(n1, n2) == ` `true` `) ` ` ` `cout << ` `"Yes"` `; ` ` ` `else` ` ` `cout << ` `"No"` `; ` ` ` ` ` `return` `0; ` `} ` |

## Java

`// An iterative java program to solve tree isomorphism problem ` ` ` `/* A binary tree node has data, pointer to left and right children */` `class` `Node ` `{ ` ` ` `int` `data; ` ` ` `Node left, right; ` ` ` ` ` `Node(` `int` `item) ` ` ` `{ ` ` ` `data = item; ` ` ` `left = right; ` ` ` `} ` `} ` ` ` `class` `BinaryTree ` `{ ` ` ` `Node root1, root2; ` ` ` ` ` `/* Given a binary tree, print its nodes in reverse level order */` ` ` `boolean` `isIsomorphic(Node n1, Node n2) ` ` ` `{ ` ` ` `// Both roots are NULL, trees isomorphic by definition ` ` ` `if` `(n1 == ` `null` `&& n2 == ` `null` `) ` ` ` `return` `true` `; ` ` ` ` ` `// Exactly one of the n1 and n2 is NULL, trees not isomorphic ` ` ` `if` `(n1 == ` `null` `|| n2 == ` `null` `) ` ` ` `return` `false` `; ` ` ` ` ` `if` `(n1.data != n2.data) ` ` ` `return` `false` `; ` ` ` ` ` `// There are two possible cases for n1 and n2 to be isomorphic ` ` ` `// Case 1: The subtrees rooted at these nodes have NOT been ` ` ` `// "Flipped". ` ` ` `// Both of these subtrees have to be isomorphic. ` ` ` `// Case 2: The subtrees rooted at these nodes have been "Flipped" ` ` ` `return` `(isIsomorphic(n1.left, n2.left) && ` ` ` `isIsomorphic(n1.right, n2.right)) ` ` ` `|| (isIsomorphic(n1.left, n2.right) && ` ` ` `isIsomorphic(n1.right, n2.left)); ` ` ` `} ` ` ` ` ` `// Driver program to test above functions ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `BinaryTree tree = ` `new` `BinaryTree(); ` ` ` ` ` `// Let us create trees shown in above diagram ` ` ` `tree.root1 = ` `new` `Node(` `1` `); ` ` ` `tree.root1.left = ` `new` `Node(` `2` `); ` ` ` `tree.root1.right = ` `new` `Node(` `3` `); ` ` ` `tree.root1.left.left = ` `new` `Node(` `4` `); ` ` ` `tree.root1.left.right = ` `new` `Node(` `5` `); ` ` ` `tree.root1.right.left = ` `new` `Node(` `6` `); ` ` ` `tree.root1.left.right.left = ` `new` `Node(` `7` `); ` ` ` `tree.root1.left.right.right = ` `new` `Node(` `8` `); ` ` ` ` ` `tree.root2 = ` `new` `Node(` `1` `); ` ` ` `tree.root2.left = ` `new` `Node(` `3` `); ` ` ` `tree.root2.right = ` `new` `Node(` `2` `); ` ` ` `tree.root2.right.left = ` `new` `Node(` `4` `); ` ` ` `tree.root2.right.right = ` `new` `Node(` `5` `); ` ` ` `tree.root2.left.right = ` `new` `Node(` `6` `); ` ` ` `tree.root2.right.right.left = ` `new` `Node(` `8` `); ` ` ` `tree.root2.right.right.right = ` `new` `Node(` `7` `); ` ` ` ` ` `if` `(tree.isIsomorphic(tree.root1, tree.root2) == ` `true` `) ` ` ` `System.out.println(` `"Yes"` `); ` ` ` `else` ` ` `System.out.println(` `"No"` `); ` ` ` `} ` `} ` ` ` `// This code has been contributed by Mayank Jaiswal ` |

## Python

`# Python program to check if two given trees are isomorphic ` ` ` `# A Binary tree node ` `class` `Node: ` ` ` `# Constructor to create the node of binary tree ` ` ` `def` `__init__(` `self` `, data): ` ` ` `self` `.data ` `=` `data ` ` ` `self` `.left ` `=` `None` ` ` `self` `.right ` `=` `None` ` ` `# Check if the binary tree is isomorphic or not ` `def` `isIsomorphic(n1, n2): ` ` ` ` ` `# Both roots are None, trees isomorphic by definition ` ` ` `if` `n1 ` `is` `None` `and` `n2 ` `is` `None` `: ` ` ` `return` `True` ` ` ` ` `# Exactly one of the n1 and n2 is None, trees are not ` ` ` `# isomorphic ` ` ` `if` `n1 ` `is` `None` `or` `n2 ` `is` `None` `: ` ` ` `return` `False` ` ` ` ` `if` `n1.data !` `=` `n2.data : ` ` ` `return` `False` ` ` `# There are two possible cases for n1 and n2 to be isomorphic ` ` ` `# Case 1: The subtrees rooted at these nodes have NOT ` ` ` `# been "Flipped". ` ` ` `# Both of these subtrees have to be isomorphic, hence the && ` ` ` `# Case 2: The subtrees rooted at these nodes have ` ` ` `# been "Flipped" ` ` ` `return` `((isIsomorphic(n1.left, n2.left)` `and` ` ` `isIsomorphic(n1.right, n2.right)) ` `or` ` ` `(isIsomorphic(n1.left, n2.right) ` `and` ` ` `isIsomorphic(n1.right, n2.left)) ` ` ` `) ` ` ` ` ` `# Driver program to test above function ` `n1 ` `=` `Node(` `1` `) ` `n1.left ` `=` `Node(` `2` `) ` `n1.right ` `=` `Node(` `3` `) ` `n1.left.left ` `=` `Node(` `4` `) ` `n1.left.right ` `=` `Node(` `5` `) ` `n1.right.left ` `=` `Node(` `6` `) ` `n1.left.right.left ` `=` `Node(` `7` `) ` `n1.left.right.right ` `=` `Node(` `8` `) ` ` ` `n2 ` `=` `Node(` `1` `) ` `n2.left ` `=` `Node(` `3` `) ` `n2.right ` `=` `Node(` `2` `) ` `n2.right.left ` `=` `Node(` `4` `) ` `n2.right.right ` `=` `Node(` `5` `) ` `n2.left.right ` `=` `Node(` `6` `) ` `n2.right.right.left ` `=` `Node(` `8` `) ` `n2.right.right.right ` `=` `Node(` `7` `) ` ` ` `print` `"Yes"` `if` `(isIsomorphic(n1, n2) ` `=` `=` `True` `) ` `else` `"No"` ` ` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) ` |

## C#

`using` `System; ` ` ` `// An iterative C# program to solve tree isomorphism problem ` ` ` `/* A binary tree node has data, pointer to left and right children */` `public` `class` `Node ` `{ ` ` ` `public` `int` `data; ` ` ` `public` `Node left, right; ` ` ` ` ` `public` `Node(` `int` `item) ` ` ` `{ ` ` ` `data = item; ` ` ` `left = right; ` ` ` `} ` `} ` ` ` `public` `class` `BinaryTree ` `{ ` ` ` `public` `Node root1, root2; ` ` ` ` ` `/* Given a binary tree, print its nodes in reverse level order */` ` ` `public` `virtual` `bool` `isIsomorphic(Node n1, Node n2) ` ` ` `{ ` ` ` `// Both roots are NULL, trees isomorphic by definition ` ` ` `if` `(n1 == ` `null` `&& n2 == ` `null` `) ` ` ` `{ ` ` ` `return` `true` `; ` ` ` `} ` ` ` ` ` `// Exactly one of the n1 and n2 is NULL, trees not isomorphic ` ` ` `if` `(n1 == ` `null` `|| n2 == ` `null` `) ` ` ` `{ ` ` ` `return` `false` `; ` ` ` `} ` ` ` ` ` `if` `(n1.data != n2.data) ` ` ` `{ ` ` ` `return` `false` `; ` ` ` `} ` ` ` ` ` `// There are two possible cases for n1 and n2 to be isomorphic ` ` ` `// Case 1: The subtrees rooted at these nodes have NOT been ` ` ` `// "Flipped". ` ` ` `// Both of these subtrees have to be isomorphic. ` ` ` `// Case 2: The subtrees rooted at these nodes have been "Flipped" ` ` ` `return` `(isIsomorphic(n1.left, n2.left) ` ` ` `&& isIsomorphic(n1.right, n2.right)) ` ` ` `|| (isIsomorphic(n1.left, n2.right) ` ` ` `&& isIsomorphic(n1.right, n2.left)); ` ` ` `} ` ` ` ` ` `// Driver program to test above functions ` ` ` `public` `static` `void` `Main(` `string` `[] args) ` ` ` `{ ` ` ` `BinaryTree tree = ` `new` `BinaryTree(); ` ` ` ` ` `// Let us create trees shown in above diagram ` ` ` `tree.root1 = ` `new` `Node(1); ` ` ` `tree.root1.left = ` `new` `Node(2); ` ` ` `tree.root1.right = ` `new` `Node(3); ` ` ` `tree.root1.left.left = ` `new` `Node(4); ` ` ` `tree.root1.left.right = ` `new` `Node(5); ` ` ` `tree.root1.right.left = ` `new` `Node(6); ` ` ` `tree.root1.left.right.left = ` `new` `Node(7); ` ` ` `tree.root1.left.right.right = ` `new` `Node(8); ` ` ` ` ` `tree.root2 = ` `new` `Node(1); ` ` ` `tree.root2.left = ` `new` `Node(3); ` ` ` `tree.root2.right = ` `new` `Node(2); ` ` ` `tree.root2.right.left = ` `new` `Node(4); ` ` ` `tree.root2.right.right = ` `new` `Node(5); ` ` ` `tree.root2.left.right = ` `new` `Node(6); ` ` ` `tree.root2.right.right.left = ` `new` `Node(8); ` ` ` `tree.root2.right.right.right = ` `new` `Node(7); ` ` ` ` ` `if` `(tree.isIsomorphic(tree.root1, tree.root2) == ` `true` `) ` ` ` `{ ` ` ` `Console.WriteLine(` `"Yes"` `); ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` `Console.WriteLine(` `"No"` `); ` ` ` `} ` ` ` `} ` `} ` ` ` `// This code is contributed by Shrikant13 ` |

Output:

Yes

**Time Complexity:** The above solution does a traversal of both trees. So time complexity is O(min(m,n)*2) or O(min(m,n)) where m and n are number of nodes in given trees.

This article is contributed by **Ciphe**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.