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)

Output:

Yes

**Time Complexity:** The above solution does a traversal of both trees. So time complexity is O(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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