Find All Duplicate Subtrees

Last Updated : 15 Jul, 2022

Given a binary tree, find all duplicate subtrees. For each duplicate subtree, we only need to return the root node of any one of them. Two trees are duplicates if they have the same structure with the same node values.

Examples:

Input :
       1
      / \
     2   3
    /   / \
   4   2   4
      /
     4

Output : 
   2           
  /    and    4
 4
Explanation: Above Trees are two duplicate subtrees. 
Therefore, you need to return above trees root in the 
form of a list.

The idea is to use hashing. We store inorder traversals of subtrees in a hash. Since simple inorder traversal cannot uniquely identify a tree, we use symbols like ‘(‘ and ‘)’ to represent NULL nodes.

We pass an Unordered Map in C++ as an argument to the helper function which recursively calculates inorder string and increases its count in map. If any string gets repeated, then it will imply duplication of the subtree rooted at that node so push that node in the Final result and return the vector of these nodes.

Implementation:

C++

// C++ program to find averages of all levels
// in a binary tree.
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data, pointer to
left child and a pointer to right child */
struct Node {
    int data;
    struct Node* left, *right;
};
 
string inorder(Node* node, unordered_map<string, int>& m)
{
    if (!node)
        return "";
 
    string str = "(";
    str += inorder(node->left, m);
    str += to_string(node->data);
    str += inorder(node->right, m);
    str += ")";
 
    // Subtree already present (Note that we use
    // unordered_map instead of unordered_set
    // because we want to print multiple duplicates
    // only once, consider example of 4 in above
    // subtree, it should be printed only once.
    if (m[str] == 1)
        cout << node->data << " ";
 
    m[str]++;
 
    return str;
}
 
// Wrapper over inorder()
void printAllDups(Node* root)
{
    unordered_map<string, int> m;
    inorder(root, m);
}
 
/* Helper function that allocates a
new node with the given data and
NULL left and right pointers. */
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Driver code
int main()
{
    Node* root = NULL;
    root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(2);
    root->right->left->left = newNode(4);
    root->right->right = newNode(4);
    printAllDups(root);
    return 0;
}

Java

// A java program to find all duplicate subtrees 
// in a binary tree.
import java.util.HashMap;
public class Duplicate_subtress {
 
    /* A binary tree node has data, pointer to
    left child and a pointer to right child */
    static HashMap<String, Integer> m;
    static class Node {
        int data;
        Node left;
        Node right;
        Node(int data){
            this.data = data;
            left = null;
            right = null;
        }
    }
    static String inorder(Node node)
    {
        if (node == null)
            return "";
      
        String str = "(";
        str += inorder(node.left);
        str += Integer.toString(node.data);
        str += inorder(node.right);
        str += ")";
      
        // Subtree already present (Note that we use
        // HashMap instead of HashSet
        // because we want to print multiple duplicates
        // only once, consider example of 4 in above
        // subtree, it should be printed only once.       
        if (m.get(str) != null && m.get(str)==1 )
            System.out.print( node.data + " ");
      
        if (m.containsKey(str))
            m.put(str, m.get(str) + 1);
        else
            m.put(str, 1);
         
        
        return str;
    }
      
    // Wrapper over inorder()
    static void printAllDups(Node root)
    {
        m = new HashMap<>();
        inorder(root);
    } 
    // Driver code
    public static void main(String args[])
    {
        Node root = null;
        root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.right.left = new Node(2);
        root.right.left.left = new Node(4);
        root.right.right = new Node(4);
        printAllDups(root);
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Python3 program to find averages of 
# all levels in a binary tree. 
 
# Helper function that allocates a 
# new node with the given data and 
# None left and right pointers. 
class newNode:
    def __init__(self, data):
        self.data = data 
        self.left = self.right = None
 
def inorder(node, m):
    if (not node): 
        return "" 
 
    Str = "("
    Str += inorder(node.left, m) 
    Str += str(node.data) 
    Str += inorder(node.right, m) 
    Str += ")"
 
    # Subtree already present (Note that 
    # we use unordered_map instead of 
    # unordered_set because we want to print
    # multiple duplicates only once, consider 
    # example of 4 in above subtree, it 
    # should be printed only once. 
    if (Str in m and m[Str] == 1): 
        print(node.data, end = " ") 
    if Str in m:
        m[Str] += 1
    else:
        m[Str] = 1
 
    return Str
 
# Wrapper over inorder() 
def printAllDups(root):
    m = {} 
    inorder(root, m)
 
# Driver code 
if __name__ == '__main__':
    root = None
    root = newNode(1) 
    root.left = newNode(2) 
    root.right = newNode(3) 
    root.left.left = newNode(4) 
    root.right.left = newNode(2) 
    root.right.left.left = newNode(4) 
    root.right.right = newNode(4) 
    printAllDups(root) 
 
# This code is contributed by PranchalK

C#

// A C# program to find all duplicate subtrees 
// in a binary tree.
using System;
using System.Collections.Generic;
 
class GFG 
{
 
    /* A binary tree node has data, pointer to
    left child and a pointer to right child */
    static Dictionary<String, 
                      int> m = new Dictionary<String, 
                                              int>();
    public class Node 
    {
        public int data;
        public Node left;
        public Node right;
        public Node(int data)
        {
            this.data = data;
            left = null;
            right = null;
        }
    }
     
    static String inorder(Node node)
    {
        if (node == null)
            return "";
     
        String str = "(";
        str += inorder(node.left);
        str += (node.data).ToString();
        str += inorder(node.right);
        str += ")";
     
        // Subtree already present (Note that we use
        // HashMap instead of HashSet
        // because we want to print multiple duplicates
        // only once, consider example of 4 in above
        // subtree, it should be printed only once.     
        if (m.ContainsKey(str) && m[str] == 1 )
            Console.Write(node.data + " ");
     
        if (m.ContainsKey(str))
            m[str] = m[str] + 1;
        else
            m.Add(str, 1);
         
        return str;
    }
     
    // Wrapper over inorder()
    static void printAllDups(Node root)
    {
        m = new Dictionary<String, int>();
        inorder(root);
    } 
     
    // Driver code
    public static void Main(String []args)
    {
        Node root = null;
        root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.right.left = new Node(2);
        root.right.left.left = new Node(4);
        root.right.right = new Node(4);
        printAllDups(root);
    }
}
 
// This code is contributed by Princi Singh

Javascript

<script>
    // A javascript program to find all duplicate subtrees
    // in a binary tree.
     
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
     
    /* A binary tree node has data, pointer to
    left child and a pointer to right child */
    let m;
     
    function inorder(node)
    {
        if (node == null)
            return "";
       
        let str = "(";
        str += inorder(node.left);
        str += toString(node.data);
        str += inorder(node.right);
        str += ")";
       
        // Subtree already present (Note that we use
        // HashMap instead of HashSet
        // because we want to print multiple duplicates
        // only once, consider example of 4 in above
        // subtree, it should be printed only once.      
        if (m.get(str) != null && m.get(str)==1 )
            document.write( node.data + " ");
       
        if (m.has(str))
            m.set(str, m.get(str) + 1);
        else
            m.set(str, 1);
          
         
        return str;
    }
       
    // Wrapper over inorder()
    function printAllDups(root)
    {
        m = new Map();
        inorder(root);
    }
     
    let root = null;
    root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.right.left = new Node(2);
    root.right.left.left = new Node(4);
    root.right.right = new Node(4);
    printAllDups(root);
    
   // This code is contributed by suresh07.
</script>

Output

4 2

Time Complexity: O(N^2) Since string copying takes O(n) extra time.
Auxiliary Space: O(N^2) Since we are hashing a string for each node and length of this string can be of the order N.