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.

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