Find All Duplicate Subtrees - GeeksforGeeks

20 Nov, 2021
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.  
C++
Java
Python3
C#
Javascript
// 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 toleft 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 anew node with the given data andNULL left and right pointers. */Node* newNode(int data){    Node* temp = new Node;    temp->data = data;    temp->left = temp->right = NULL;    return temp;} // Driver codeint 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;}
// 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 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 codeif __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
// 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
<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.If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
PranchalKatiyar
princi singh
rohitsingh07052
helloween123
suresh07
neminbshah
ankurk1947
cpp-unordered_map
Hash
Tree
Hash
Tree
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Given an array A[] and a number x, check for pair in A[] with sum as x (aka Two Sum)
Internal Working of HashMap in Java
Hashing | Set 1 (Introduction)
Hashing | Set 3 (Open Addressing)
Hashing | Set 2 (Separate Chaining)
Tree Traversals (Inorder, Preorder and Postorder)
AVL Tree | Set 1 (Insertion)
Binary Tree | Set 1 (Introduction)
Level Order Binary Tree Traversal
Binary Tree | Set 3 (Types of Binary Tree)