Check if a Binary Tree (not BST) has duplicate values
Check if a Binary Tree (not BST) has duplicate values
To check if a binary tree has duplicate values, you can follow these steps:
- Create a set to store the values that have been encountered so far.
- 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.
- 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++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node* left;
struct Node* right;
};
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 (root == NULL)
return false ;
if (s.find(root->data) != s.end())
return true ;
s.insert(root->data);
return checkDupUtil(root->left, s) ||
checkDupUtil(root->right, s);
}
bool checkDup( struct Node* root)
{
unordered_set< int > s;
return checkDupUtil(root, s);
}
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
import java.util.HashSet;
public class CheckDuplicateValues {
public static boolean checkDupUtil(Node root, HashSet<Integer> s)
{
if (root == null )
return false ;
if (s.contains(root.data))
return true ;
s.add(root.data);
return checkDupUtil(root.left, s) || checkDupUtil(root.right, s);
}
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" );
}
}
class Node {
int data;
Node left,right;
Node( int data)
{
this .data=data;
}
};
|
Python
class newNode:
def __init__( self , key):
self .data = key
self .left = None
self .right = None
def checkDupUtil( root, s) :
if (root = = None ) :
return False
if root.data in s:
return True
s.add(root.data)
return checkDupUtil(root.left, s) or checkDupUtil(root.right, s)
def checkDup( root) :
s = set ()
return checkDupUtil(root, s)
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" )
|
C#
using System;
using System.Collections;
using System.Collections.Generic;
class CheckDuplicateValues
{
public static Boolean checkDupUtil(Node root, HashSet< int > s)
{
if (root == null )
return false ;
if (s.Contains(root.data))
return true ;
s.Add(root.data);
return checkDupUtil(root.left, s) ||
checkDupUtil(root.right, s);
}
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" );
}
}
public class Node
{
public int data;
public Node left, right;
public Node( int data)
{
this .data = data;
}
};
|
Javascript
<script>
class Node
{
constructor(data) {
this .left = null ;
this .right = null ;
this .data = data;
}
}
function checkDupUtil(root, s)
{
if (root == null )
return false ;
if (s.has(root.data))
return true ;
s.add(root.data);
return checkDupUtil(root.left, s) ||
checkDupUtil(root.right, s);
}
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>
|
Time complexity: O(n), The time complexity of the above code is O(n) as we traverse the tree once and insert each node into the unordered set.
Auxiliary Space: O(n),The space complexity of the above code is O(n) as we store all the nodes into the unordered set.
EXAMPLE 2 :
C++
#include <iostream>
#include <unordered_set>
using namespace std;
class Node {
public :
int data;
Node* left;
Node* right;
Node( int data)
{
this ->data = data;
this ->left = NULL;
this ->right = NULL;
}
};
class BinaryTree {
public :
Node* root;
bool hasDuplicateValues()
{
unordered_set< int > values;
return hasDuplicateValues(root, values);
}
private :
bool hasDuplicateValues(Node* focusNode,
unordered_set< int >& values)
{
if (focusNode == NULL) {
return false ;
}
if (values.count(focusNode->data) > 0) {
return true ;
}
values.insert(focusNode->data);
return hasDuplicateValues(focusNode->left, values)
|| hasDuplicateValues(focusNode->right,
values);
}
};
int main()
{
BinaryTree tree;
tree.root = new Node(1);
tree.root->left = new Node(2);
tree.root->right = new Node(2);
cout << boolalpha << tree.hasDuplicateValues()
<< endl;
tree.root->right = new Node(3);
cout << boolalpha << tree.hasDuplicateValues()
<< endl;
return 0;
}
|
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());
tree.root.right = new Node( 3 );
System.out.println(tree.hasDuplicateValues());
}
}
|
Python3
class Node:
def __init__( self , data):
self .data = data
self .left = None
self .right = None
class BinaryTree:
def __init__( self ):
self .root = None
def hasDuplicateValues( self ):
values = set ()
return self .hasDuplicateValuesRecursive( self .root, values)
def hasDuplicateValuesRecursive( self , focusNode, values):
if focusNode is None :
return False
if focusNode.data in values:
return True
values.add(focusNode.data)
return self .hasDuplicateValuesRecursive(focusNode.left, values) or self .hasDuplicateValuesRecursive(focusNode.right, values)
def main():
tree = BinaryTree()
tree.root = Node( 1 )
tree.root.left = Node( 2 )
tree.root.right = Node( 2 )
print (tree.hasDuplicateValues())
tree.root.right = Node( 3 )
print (tree.hasDuplicateValues())
if __name__ = = '__main__' :
main()
|
C#
using System;
using System.Collections.Generic;
public class Node
{
public int data;
public Node left;
public Node right;
public Node( int data)
{
this .data = data;
}
}
public class BinaryTree
{
public Node root;
public bool HasDuplicateValues()
{
HashSet< int > values = new HashSet< int >();
return HasDuplicateValues(root, values);
}
private bool HasDuplicateValues(Node focusNode, HashSet< int > 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 MainClass
{
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);
Console.WriteLine(tree.HasDuplicateValues());
tree.root.right = new Node(3);
Console.WriteLine(tree.HasDuplicateValues());
}
}
|
Javascript
class Node {
constructor(data) {
this .data = data;
this .left = null ;
this .right = null ;
}
}
class BinaryTree {
constructor() {
this .root = null ;
}
hasDuplicateValues() {
const values = new Set();
return this ._hasDuplicateValues( this .root, values);
}
_hasDuplicateValues(focusNode, values) {
if (focusNode === null ) {
return false ;
}
if (values.has(focusNode.data)) {
return true ;
}
values.add(focusNode.data);
return (
this ._hasDuplicateValues(focusNode.left, values) ||
this ._hasDuplicateValues(focusNode.right, values)
);
}
}
const tree = new BinaryTree();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(2);
console.log(tree.hasDuplicateValues());
tree.root.right = new Node(3);
console.log(tree.hasDuplicateValues());
|
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).
Last Updated :
02 Mar, 2023
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