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

Last Updated : 03 May, 2024

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 the 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++

// 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
        // Append random char in between the value to
        // differentiate from 11,1 and 1,11
        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 7 (3+4 accounting for special
        // chars appended) as it has two marker nodes.
        if (s.length() > 7 && 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

Python

# Python program to find if there is a duplicate
# sub-tree of size 2 or more.

# A binary tree node has data,
# pointer to left child and a
# pointer to right child


class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None


def isSame(T, S):
    if T is None and S is None:
        return True
    if T is None or S is None:
        return False
    if T.data == S.data and isSame(T.left, S.left) and isSame(T.right, S.right):
        return True
    return False


def isSubtree(T, S):
    if T is None or S is None:
        return False
    if T == S:
        return False
    return isSame(T, S) or isSubtree(T.left, S) or isSubtree(T.right, S)


def findDuplicates(T, S):
    if T is None or S is None:
        return False
    if S.left is None and S.right is None:
        return False
    if T != S:
        if isSubtree(T, S):
            return True
    left = findDuplicates(T, S.left)
    if left:
        return True
    right = findDuplicates(T, S.right)
    return right


def dupSub(root):
    if root is None:
        return 0
    return 1 if findDuplicates(root, root) else 0


# Driver code
root = Node('A')
root.left = Node('B')
root.right = Node('C')
root.left.left = Node('D')
root.left.right = Node('E')
root.right.right = Node('B')
root.right.right.right = Node('E')
root.right.right.left = Node('D')

if dupSub(root):
    print("Yes")
else:
    print("No")

// 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

Javascript

// Javascript program to find if there is a duplicate
// sub-tree of size 2 or more.
    
    let MARKER = '$';
    
    // A binary tree Node has data,
// pointer to left child
// and a pointer to right child
    class Node {
        constructor(data)
        {
            this.data=data;
        }
    }
    
    // This function returns empty string if tree
    // contains a duplicate subtree of size 2 or more.
    function dupSubUtil(root,subtrees)
    {
        let s = "";
     
        // If current node is NULL, return marker
        if (root == null)
            return s + MARKER;
     
        // If left subtree has a duplicate subtree.
        let lStr = dupSubUtil(root.left,subtrees);
        if (lStr==(s))
            return s;
     
        // Do same for right subtree
        let rStr = dupSubUtil(root.right,subtrees);
        if (rStr==(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.has(s))
            return "";
     
        subtrees.add(s);
        return s;
    }
    
    //Function to find if the Binary Tree contains duplicate
    //subtrees of size 2 or more
    function dupSub(root)
    {
            let subtrees=new Set();
            
            return dupSubUtil(root,subtrees);
    }
    
    let 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');
    let str = dupSub(root);
    if(str==(""))
        console.log(" Yes ");
     else   
        console.log(" No ");
    

// This code is contributed by unknown2108

Output

Yes

Time Complexity: O(n)
Auxiliary Space: O(n)

Nishant Singh

Improve

Check whether a given binary tree is perfect or not

Check if two trees are Mirror

Types of Binary Tree

Some important Binary tree traversals

Easy problems on Binary Tree

Intermediate problems on Binary Tree

Hard problems on Binary Tree

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

Method 1 :

[Method 2 ]( Efficient solution )

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