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Check if a Binary Tree (not BST) has duplicate values

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  • Difficulty Level : Basic
  • Last Updated : 12 Jan, 2023
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Check if a Binary Tree (not BST) has duplicate values

To check if a binary tree has duplicate values, you can follow these steps:

  1. Create a set to store the values that have been encountered so far.
  2. Start a traversal of the binary tree. For each node, check if its value is in the set. If it is, return true to indicate that the tree contains duplicate values. If it is not, add the value to the set and continue the traversal.
  3. When the traversal is complete, return false to indicate that no duplicate values were found.

Examples: 

Input : Root of below tree
         1
       /   \
      2     3
             \
              2
Output : Yes
Explanation : The duplicate value is 2.

Input : Root of below tree
         1
       /   \
     20     3
             \
              4
Output : No
Explanation : There are no duplicates.

A simple solution is to store inorder traversal of given binary tree in an array. Then check if array has duplicates or not. We can avoid the use of array and solve the problem in O(n) time. The idea is to use hashing. We traverse the given tree, for every node, we check if it already exists in hash table. If exists, we return true (found duplicate). If it does not exist, we insert into hash table.

Implementation:

C++




// C++ Program to check duplicates
// in 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;
    struct Node* right;
};
 
// Helper function that allocates
// a new Node with the given data
// and NULL left and right pointers.
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
 
bool checkDupUtil(Node* root, unordered_set<int> &s)
{
    // If tree is empty, there are no
    // duplicates.
    if (root == NULL)
       return false;
 
    // If current node's data is already present.
    if (s.find(root->data) != s.end())
       return true;
 
    // Insert current node
    s.insert(root->data);
     
    // Recursively check in left and right
    // subtrees.
    return checkDupUtil(root->left, s) ||
           checkDupUtil(root->right, s);
}
 
// To check if tree has duplicates
bool checkDup(struct Node* root)
{
    unordered_set<int> s;
    return checkDupUtil(root, s);
}
 
// Driver program to test above functions
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(2);
    root->left->left = newNode(3);
    if (checkDup(root))
        printf("Yes");
    else
        printf("No");
 
    return 0;
}

Java




// Java Program to check duplicates
// in Binary Tree
import java.util.HashSet;
public class CheckDuplicateValues {
 
    //Function that used HashSet to find presence of duplicate nodes
    public static boolean checkDupUtil(Node root, HashSet<Integer> s)
    {
        // If tree is empty, there are no
        // duplicates. 
        if (root == null)
            return false;
   
        // If current node's data is already present.
        if (s.contains(root.data))
            return true;
   
        // Insert current node
        s.add(root.data);
       
        // Recursively check in left and right
        // subtrees.
        return checkDupUtil(root.left, s) || checkDupUtil(root.right, s);
    }
   
    // To check if tree has duplicates
    public static boolean checkDup(Node root)
    {
        HashSet<Integer> s=new HashSet<>();
        return checkDupUtil(root, s);
    }
 
    public static void main(String args[]) {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(2);
        root.left.left = new Node(3);
        if (checkDup(root))
            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




""" Program to check duplicates
# in Binary Tree """
 
# Helper function that allocates a new
# node with the given data and None
# left and right pointers.                                
class newNode:
 
    # Construct to create a new node
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
def checkDupUtil( root, s) :
 
    # If tree is empty, there are no
    # duplicates.
    if (root == None) :
        return False
 
    # If current node's data is already present.
    if root.data in s:
        return True
 
    # Insert current node
    s.add(root.data)
     
    # Recursively check in left and right
    # subtrees.
    return checkDupUtil(root.left, s) or checkDupUtil(root.right, s)
 
 
# To check if tree has duplicates
def checkDup( root) :
 
    s=set()
    return checkDupUtil(root, s)
 
 
# Driver Code
if __name__ == '__main__':
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(2)
    root.left.left = newNode(3)
    if (checkDup(root)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#




// C# Program to check duplicates
// in Binary Tree
using System;
using System.Collections;
using System.Collections.Generic;
 
class CheckDuplicateValues
{
 
    //Function that used HashSet to
    // find presence of duplicate nodes
    public static Boolean checkDupUtil(Node root, HashSet<int> s)
    {
        // If tree is empty, there are no
        // duplicates.
        if (root == null)
            return false;
     
        // If current node's data is already present.
        if (s.Contains(root.data))
            return true;
     
        // Insert current node
        s.Add(root.data);
         
        // Recursively check in left and right
        // subtrees.
        return checkDupUtil(root.left, s) ||
                checkDupUtil(root.right, s);
    }
     
    // To check if tree has duplicates
    public static Boolean checkDup(Node root)
    {
        HashSet<int> s = new HashSet<int>();
        return checkDupUtil(root, s);
    }
 
    public static void Main(String []args)
    {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(2);
        root.left.left = new Node(3);
        if (checkDup(root))
            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 Arnab Kundu

Javascript




<script>
 
    // JavaScript Program to check duplicates in Binary Tree
     
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
 
    // Function that used HashSet to find
    // presence of duplicate nodes
    function checkDupUtil(root, s)
    {
        // If tree is empty, there are no
        // duplicates. 
        if (root == null)
            return false;
     
        // If current node's data is already present.
        if (s.has(root.data))
            return true;
     
        // Insert current node
        s.add(root.data);
         
        // Recursively check in left and right
        // subtrees.
        return checkDupUtil(root.left, s) ||
        checkDupUtil(root.right, s);
    }
     
    // To check if tree has duplicates
    function checkDup(root)
    {
        let s = new Set();
        return checkDupUtil(root, s);
    }
     
    let root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(2);
    root.left.left = new Node(3);
    if (checkDup(root))
      document.write("Yes");
    else
      document.write("No");
     
</script>

Output

Yes

EXAMPLE 2 :

Java




import java.util.HashSet;
 
class Node {
  int data;
  Node left;
  Node right;
 
  public Node(int data) {
    this.data = data;
  }
}
 
class BinaryTree {
  Node root;
 
  public boolean hasDuplicateValues() {
    HashSet<Integer> values = new HashSet<>();
    return hasDuplicateValues(root, values);
  }
 
  private boolean hasDuplicateValues(Node focusNode, HashSet<Integer> values) {
    if (focusNode == null) {
      return false;
    }
 
    if (values.contains(focusNode.data)) {
      return true;
    }
 
    values.add(focusNode.data);
    return hasDuplicateValues(focusNode.left, values) || hasDuplicateValues(focusNode.right, values);
  }
}
 
public class Main {
  public static void main(String[] args) {
    BinaryTree tree = new BinaryTree();
 
    tree.root = new Node(1);
    tree.root.left = new Node(2);
    tree.root.right = new Node(2);
 
    System.out.println(tree.hasDuplicateValues()); // prints "true"
 
    tree.root.right = new Node(3);
 
    System.out.println(tree.hasDuplicateValues()); // prints "false"
  }
}

Output

true
false

Complexity analysis:

The hasDuplicateValues method has a time complexity of O(n), where n is the number of nodes in the binary tree. This is because the method visits each node exactly once in order to check if it has already been seen before.

The space complexity of the hasDuplicateValues method is O(m), where m is the number of unique values in the binary tree. The HashSet values stores all the unique values that have been seen so far, and its size can at most be equal to the number of unique values in the binary tree. This is because if there are no duplicates in the tree, the size of the HashSet will be equal to the number of nodes in the tree, which is n.
Otherwise, if there are duplicates, the size of the set will be less than the number of nodes, hence m<=n.

In the worst-case scenario, where all the values in the tree are different, the space complexity of the method would be O(n). In the best-case scenario, where all the values in the tree are the same, the space complexity of the method would be O(1).


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