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

Tree
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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Output:

Yes

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