C Program to Check if Binary Tree is Subtree of Another Tree

This is a C Program program to check whether a binary tree is subtree of another tree. Given two binary trees, check if the first tree is subtree of the second one. A subtree of a tree T is a tree S consisting of a node in T and all of its descendants in T. The subtree corresponding to the root node is the entire tree; the subtree corresponding to any other node is called a proper subtree.

Here is source code of the C Program to Check Whether an Input Binary Tree is the Sub Tree of the Binary Tree. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

```
#include <stdio.h>
```
```
#include <stdlib.h>
```
```
 
```

/* A binary tree node has data, left child and right child */

```
struct node {
```
```
    int data;
```
```
    struct node* left;
```
```
    struct node* right;
```
```
};
```
```
 
```

/* A utility function to check whether trees with roots as root1 and

```
 root2 are identical or not */
```

bool areIdentical(struct node * root1, struct node *root2) {

```
    /* base cases */
```
```
    if (root1 == NULL && root2 == NULL)
```
```
        return true;
```
```
 
```
```
    if (root1 == NULL || root2 == NULL)
```
```
        return false;
```
```
 
```

    /* Check if the data of both roots is same and data of left and right

```
     subtrees are also same */
```
```
    return (root1->data == root2->data
```

            && areIdentical(root1->left, root2->left) && areIdentical(

            root1->right, root2->right));

```
}
```
```
 
```

/* This function returns true if S is a subtree of T, otherwise false */

bool isSubtree(struct node *T, struct node *S) {

```
    /* base cases */
```
```
    if (S == NULL)
```
```
        return true;
```
```
 
```
```
    if (T == NULL)
```
```
        return false;
```
```
 
```

    /* Check the tree with root as current node */

```
    if (areIdentical(T, S))
```
```
        return true;
```
```
 
```

    /* If the tree with root as current node doesn't match then

     try left and right subtrees one by one */

    return isSubtree(T->left, S) || isSubtree(T->right, S);

```
}
```
```
 
```

/* Helper function that allocates a new node with the given data

```
 and NULL left and right pointers. */
```
```
struct node* newNode(int data) {
```

    struct node* node = (struct node*) malloc(sizeof(struct node));

```
    node->data = data;
```
```
    node->left = NULL;
```
```
    node->right = NULL;
```
```
    return (node);
```
```
}
```
```
 
```

/* Driver program to test above function */

```
int main() {
```
```
    /* Construct the following tree
```
```
       26
```
```
     /   \
```
```
    10     3
```
```
  /    \     \
```
```
 4      6      3
```
```
  \
```
```
   30
```
```
     */
```
```
    struct node *T = newNode(26);
```
```
    T->right = newNode(3);
```
```
    T->right->right = newNode(3);
```
```
    T->left = newNode(10);
```
```
    T->left->left = newNode(4);
```
```
    T->left->left->right = newNode(30);
```
```
    T->left->right = newNode(6);
```
```
 
```
```
    /* Construct the following tree
```
```
       10
```
```
     /    \
```
```
    4      6
```
```
     \
```
```
      30
```
```
     */
```
```
    struct node *S = newNode(10);
```
```
    S->right = newNode(6);
```
```
    S->left = newNode(4);
```
```
    S->left->right = newNode(30);
```
```
 
```
```
    if (isSubtree(T, S))
```

        printf("Tree S is subtree of tree T");

```
    else
```

        printf("Tree S is not a subtree of tree T");

```
 
```
```
    getchar();
```
```
    return 0;
```
```
}
```

Output:

$ gcc SubTree.c
$ ./a.out
 
Tree S is subtree of tree T

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