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 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 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 |
Python3
# Python3 program to find if there is# a duplicate sub-tree of size 2 or more# Separator nodeMARKER = '$'# Structure for a binary tree nodeclass Node: def __init__(self, x): self.key = x self.left = None self.right = Nonesubtrees = {}# This function returns empty if tree# contains a duplicate subtree of size# 2 or more.def dupSubUtil(root): global subtrees s = "" # If current node is None, return marker if (root == None): return s + MARKER # If left subtree has a duplicate subtree. lStr = dupSubUtil(root.left) if (s in lStr): return s # Do same for right subtree rStr = dupSubUtil(root.right) if (s in rStr): return s # Serialize current subtree s = s + root.key + lStr + rStr # If current subtree already exists in hash # table. [Note that size of a serialized tree # with single node is 3 as it has two marker # nodes. if (len(s) > 3 and s in subtrees): return "" subtrees[s] = 1 return s# Driver codeif __name__ == '__main__': root = Node('A') root.left = Node('B') root.right = Node('C') root.left.left = Node('D') root.left.right = Node('E') root.right.right = Node('B') root.right.right.right = Node('E') root.right.right.left= Node('D') str = dupSubUtil(root) if "" in str: print(" Yes ") else: print(" No ")# This code is contributed by mohit kumar 29 |
C#
// C# program to find if there is a duplicate // sub-tree of size 2 or more. using System; using System.Collections.Generic; class GFG { static char MARKER = '$'; // 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 |
Output:
Yes
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