Check if a Binary Tree contains duplicate subtrees of size 2 or more


Given a Binary Tree, check whether the Binary tree contains a duplicate sub-tree of size 2 or more.
Note : Two same leaf nodes are not considered as subtree size of a leaf node is one.

Input :  Binary Tree 
             /    \ 
           B        C
         /   \       \    
        D     E       B     
                     /  \    
                    D    E
Output : Yes

Asked in : Google Interview

Tree with duplicate Sub-Tree [ highlight by blue color ellipse ]

[ Method 1]
A simple solution is that, we pick every node of tree and try to find is any sub-tree of given tree is present in tree which is identical with that sub-tree. Here we can use below post to find if a subtree is present anywhere else in tree.
Check if a binary tree is subtree of another binary tree

[Method 2 ]( Efficient solution )
An Efficient solution based on tree serialization and hashing. The idea is to serialize subtrees as strings and store the strings in hash table. Once we find a serialized tree (which is not a leaf) already existing in hash-table, we return true.
Below c++ implementation of above idea.

// C++ program to find if there is a duplicate
// sub-tree of size 2 or more.
using namespace std;

// Separator node
const char MARKER = '$';

// Structure for a binary tree node
struct Node
    char key;
    Node *left, *right;

// A utility function to create a new node
Node* newNode(char key)
    Node* node = new Node;
    node->key = key;
    node->left = node->right = NULL;
    return node;

unordered_set<string> subtrees;

// This function returns empty string if tree
// contains a duplicate subtree of size 2 or more.
string dupSubUtil(Node *root)
    string s = "";

    // If current node is NULL, return marker
    if (root == NULL)
        return s + MARKER;

    // If left subtree has a duplicate subtree.
    string lStr = dupSubUtil(root->left);
    if ( == 0)
       return s;

    // Do same for right subtree
    string rStr = dupSubUtil(root->right);
    if ( == 0)
       return s;

    // Serialize current subtree
    s = s + root->key + lStr + rStr;

    // If current subtree already exists in hash
    // table. [Note that size of a serialized tree
    // with single node is 3 as it has two marker
    // nodes.
    if (s.length() > 3 &&
        subtrees.find(s) != subtrees.end())
       return "";


    return s;

// Driver program to test above functions
int main()
    Node *root = newNode('A');
    root->left = newNode('B');
    root->right = newNode('C');
    root->left->left = newNode('D');
    root->left->right = newNode('E');
    root->right->right = newNode('B');
    root->right->right->right = newNode('E');
    root->right->right->left= newNode('D');

    string str = dupSubUtil(root);

    ("") == 0) ? cout << " Yes ":
                             cout << " No " ;
    return 0;



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