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Copy pathisomorphicBST.cpp
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77 lines (65 loc) · 2.06 KB
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// 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;
}