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 
               A
             /    \ 
           B        C
         /   \       \    
        D     E       B     
                     /  \    
                    D    E
Output : Yes

Asked in : Google Interview

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 The implementation of above idea.

C++

// C++ program to find if there is a duplicate 
// sub-tree of size 2 or more. 
#include<bits/stdc++.h> 
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 (lStr.compare(s) == 0) 
       return s; 
  
    // Do same for right subtree 
    string rStr = dupSubUtil(root->right); 
    if (rStr.compare(s) == 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 ""; 
  
    subtrees.insert(s); 
  
    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); 
  
    (str.compare("") == 0) ? cout << " Yes ": 
                             cout << " No " ; 
    return 0; 
} 

Java

// Java program to find if there is a duplicate  
// sub-tree of size 2 or more.  
import java.util.HashSet; 
public class Main {  
  
    static char MARKER = '$'; 
  
    // This function returns empty string if tree  
    // contains a duplicate subtree of size 2 or more.  
    public static String dupSubUtil(Node root, HashSet<String> subtrees)  
    {  
        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,subtrees);  
        if (lStr.equals(s))  
            return s;  
    
        // Do same for right subtree  
        String rStr = dupSubUtil(root.right,subtrees);  
        if (rStr.equals(s))  
            return s;  
    
        // Serialize current subtree  
        s = s + root.data + 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.contains(s))  
            return "";  
    
        subtrees.add(s);  
        return s;  
    }  
  
    //Function to find if the Binary Tree contains duplicate  
    //subtrees of size 2 or more 
    public static String dupSub(Node root) 
    { 
        HashSet<String> subtrees=new HashSet<>(); 
        return dupSubUtil(root,subtrees); 
    } 
  
    public static void main(String args[])  
    { 
        Node root = new Node('A');  
        root.left = new Node('B');  
        root.right = new Node('C');  
        root.left.left = new Node('D');  
        root.left.right = new Node('E');  
        root.right.right = new Node('B');  
        root.right.right.right = new Node('E');  
        root.right.right.left= new Node('D');  
        String str = dupSub(root);  
        if(str.equals("")) 
            System.out.print(" Yes "); 
        else    
            System.out.print(" No ");  
    } 
} 
  
// A binary tree Node has data,  
// pointer to left child  
// and a pointer to right child  
class Node {  
    int data;  
    Node left,right;  
    Node(int data) 
    { 
        this.data=data; 
    } 
}; 
//This code is contributed by Gaurav Tiwari 

C#

// C# program to find if there is a duplicate  
// sub-tree of size 2 or more. 
using System; 
using System.Collections.Generic; 
  
class GFG  
{  
  
    static char MARKER = '$'; 
  
    // This function returns empty string if tree  
    // contains a duplicate subtree of size 2 or more.  
    public static String dupSubUtil(Node root,  
                    HashSet<String> subtrees)  
    {  
        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,subtrees);  
        if (lStr.Equals(s))  
            return s;  
      
        // Do same for right subtree  
        String rStr = dupSubUtil(root.right,subtrees);  
        if (rStr.Equals(s))  
            return s;  
      
        // Serialize current subtree  
        s = s + root.data + 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.Contains(s))  
            return "";  
      
        subtrees.Add(s);  
        return s;  
    }  
  
    // Function to find if the Binary Tree contains   
    // duplicate subtrees of size 2 or more 
    public static String dupSub(Node root) 
    { 
        HashSet<String> subtrees = new HashSet<String>(); 
        return dupSubUtil(root,subtrees); 
    } 
  
    // Driver code 
    public static void Main(String []args)  
    { 
        Node root = new Node('A');  
        root.left = new Node('B');  
        root.right = new Node('C');  
        root.left.left = new Node('D');  
        root.left.right = new Node('E');  
        root.right.right = new Node('B');  
        root.right.right.right = new Node('E');  
        root.right.right.left= new Node('D');  
        String str = dupSub(root);  
        if(str.Equals("")) 
            Console.Write(" Yes "); 
        else
            Console.Write(" No ");  
    } 
} 
  
// A binary tree Node has data,  
// pointer to left child  
// and a pointer to right child  
public class Node  
{  
    public int data;  
    public Node left,right;  
    public Node(int data) 
    { 
        this.data = data; 
    } 
}; 
  
// This code is contributed by 29AjayKumar 

Output:

Yes

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

C#

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