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.

For example, in the following case, tree S is a subtree of tree T.

Tree 210 / \ 4 6 \ 30

Tree 126 / \ 10 3 / \ \ 4 6 3 \ 30

**Solution:** Traverse the tree T in preorder fashion. For every visited node in the traversal, see if the subtree rooted with this node is identical to S.

Following is the implementation for this.

## C

#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() { // TREE 1 /* 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); // TREE 2 /* 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 2 is subtree of Tree 1"); else printf("Tree 2 is not a subtree of Tree 1"); getchar(); return 0; }

## Java

// Java program to check if binary tree is subtree of another binary tree // A binary tree node class Node { int data; Node left, right, nextRight; Node(int item) { data = item; left = right = nextRight = null; } } class BinaryTree { Node root1,root2; /* A utility function to check whether trees with roots as root1 and root2 are identical or not */ boolean areIdentical(Node root1, 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 */ boolean isSubtree(Node T, 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); } public static void main(String args[]) { BinaryTree tree = new BinaryTree(); // TREE 1 /* Construct the following tree 26 / \ 10 3 / \ \ 4 6 3 \ 30 */ tree.root1 = new Node(26); tree.root1.right = new Node(3); tree.root1.right.right = new Node(3); tree.root1.left = new Node(10); tree.root1.left.left = new Node(4); tree.root1.left.left.right = new Node(30); tree.root1.left.right = new Node(6); // TREE 2 /* Construct the following tree 10 / \ 4 6 \ 30 */ tree.root2 = new Node(10); tree.root2.right = new Node(6); tree.root2.left = new Node(4); tree.root2.left.right = new Node(30); if (tree.isSubtree(tree.root1, tree.root2)) System.out.println("Tree 2 is subtree of Tree 1 "); else System.out.println("Tree 2 is not a subtree of Tree 1"); } } // This code has been contributed by Mayank Jaiswal

## Python

# Python program to check binary tree is a subtree of # another tree # A binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # A utility function to check whether trees with roots # as root 1 and root2 are indetical or not def areIdentical(root1, root2): # Base Case if root1 is None and root2 is None: return True if root1 is None or root2 is None: return False # Check fi the data of both roots is same and data of # left and right subtrees are also same return (root1.data == root2.data and areIdentical(root1.left , root2.left)and areIdentical(root1.right, root2.right) ) # This function returns True if S is a subtree of T, # otherwise False def isSubtree(T, S): # Base Case if S is None: return True if T is None: return True # 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 subtreee one by one return isSubtree(T.left, S) or isSubtree(T.right, S) # Driver program to test above function """ TREE 1 Construct the following tree 26 / \ 10 3 / \ \ 4 6 3 \ 30 """ T = Node(26) T.right = Node(3) T.right.right = Node(3) T.left = Node(10) T.left.left = Node(4) T.left.left.right = Node(30) T.left.right = Node(6) """ TREE 2 Construct the following tree 10 / \ 4 6 \ 30 """ S = Node(10) S.right = Node(6) S.left = Node(4) S.left.right = Node(30) if isSubtree(T, S): print "Tree 2 is subtree of Tree 1" else : print "Tree 2 is not a subtree of Tree 1" # This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Output:

Tree 2 is subtree of Tree 1

Time Complexity: Time worst case complexity of above solution is O(mn) where m and n are number of nodes in given two trees.

We can solve the above problem in O(n) time. Please refer Check if a binary tree is subtree of another binary tree | Set 2 for O(n) solution.

### Asked in: Cavisson System, Microsoft

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