# Find All Duplicate Subtrees

• Difficulty Level : Hard
• 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.```
Recommended Practice

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 ` `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& 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 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 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 m = ``new` `Dictionary();` `    ``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();` `        ``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

 ``

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.

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