Find maximum count of duplicate nodes in a Binary Search Tree

Last Updated : 16 Aug, 2021

Given a Binary Search Tree (BST) with duplicates, find the node (the most frequently occurred element) in the given BST. If the BST contains two or more such nodes, print any of them.
Note: We cannot use any extra space. (Assume that the implicit stack space incurred due to recursion does not count)
Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than or equal to the node’s key.
The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
Both the left and right subtrees must also be binary search trees.

Examples:

Input :   Given BST is                      6                  /    \                 5       7               /   \    /  \              4     5  7    7 Output : 7   Input :  Given BST is                       10                  /    \                 5       12               /   \    /  \              5     6  12    16 Output : 5 or 12  We can print any of the two value 5 or 12.

Approach:
To find the node, we need to find the Inorder Traversal of the BST because its Inorder Traversal will be in sorted order.
So, the idea is to do recursive Inorder traversal and keeping the track of the previous node. If the current node value is equal to the previous value we can increase the current count and if the current count becomes greater than the maximum count, replace the element.
Below is the implementation of the above approach:

C++

/* C++ program to find the median of BST in O(n) 
   time and O(1) space*/
 
#include <iostream>
using namespace std;
 
/* A binary search tree Node has data, pointer
   to left child and a pointer to right child */
 
struct Node {
    int val;
    struct Node *left, *right;
};
 
struct Node* newNode(int data)
{
    struct Node* temp = new Node;
    temp->val = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// cur for storing the current count of the value
// and mx for the maximum count of the element which is denoted by node
 
int cur = 1, mx = 0;
int node;
struct Node* previous = NULL;
 
// Find the inorder traversal of the BST
void inorder(struct Node* root)
{
    // If root is NULL then return
    if (root == NULL) {
        return;
    }
    inorder(root->left);
    if (previous != NULL) {
        // If the previous value is equal to the current value
        // then increase the count
        if (root->val == previous->val) {
            cur++;
        }
        // Else initialize the count by 1
        else {
            cur = 1;
        }
    }
    // If current count is greater than the max count
    // then update the mx value
    if (cur > mx) {
        mx = cur;
        node = root->val;
    }
    // Make the current Node as previous
    previous = root;
    inorder(root->right);
}
 
// Utility function
int findnode(struct Node* root)
{
    inorder(root);
    return node;
}
int main()
{
    /* Let us create following BST
                   6
                 /    \
                5       7
              /   \    /  \
             4     5  7    7 
    */
 
    struct Node* root = newNode(6);
    root->left = newNode(5);
    root->right = newNode(7);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(7);
    root->right->right = newNode(7);
 
    cout << "Node of BST is " << findnode(root) << '\n';
    return 0;
}

Java

/* Java program to find the median of BST 
in O(n) time and O(1) space*/
class GFG
{
     
/* A binary search tree Node has data, pointer
to left child and a pointer to right child */
static class Node
{
    int val;
    Node left, right;
};
 
static Node newNode(int data)
{
    Node temp = new Node();
    temp.val = data;
    temp.left = temp.right = null;
    return temp;
}
 
// cur for storing the current count 
// of the value and mx for the maximum count
// of the element which is denoted by node
static int cur = 1, mx = 0;
static int node;
static Node previous = null;
 
// Find the inorder traversal of the BST
static void inorder(Node root)
{
    // If root is null then return
    if (root == null) 
    {
        return;
    }
    inorder(root.left);
    if (previous != null) 
    {
         
        // If the previous value is equal to 
        // the current value then increase the count
        if (root.val == previous.val) 
        {
            cur++;
        }
         
        // Else initialize the count by 1
        else
        {
            cur = 1;
        }
    }
     
    // If current count is greater than the 
    // max count then update the mx value
    if (cur > mx) 
    {
        mx = cur;
        node = root.val;
    }
     
    // Make the current Node as previous
    previous = root;
    inorder(root.right);
}
 
// Utility function
static int findnode(Node root)
{
    inorder(root);
    return node;
}
 
// Java Code
public static void main(String args[])
{
    /* Let us create following BST
                6
                / \
                5     7
            / \ / \
            4     5 7 7 
    */
    Node root = newNode(6);
    root.left = newNode(5);
    root.right = newNode(7);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(7);
    root.right.right = newNode(7);
 
    System.out.println("Node of BST is " + 
                          findnode(root));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python program to find the median of BST
# in O(n) time and O(1) space
 
# A binary search tree Node has data, pointer
# to left child and a pointer to right child
class Node:
    def __init__(self):
        self.val = 0
        self.left = None
        self.right = None
 
def newNode(data: int) -> Node:
    temp = Node()
    temp.val = data
    temp.left = temp.right = None
    return temp
 
# cur for storing the current count
# of the value and mx for the maximum count
# of the element which is denoted by node
cur = 1
mx = 0
node = 0
previous = Node()
 
# Find the inorder traversal of the BST
def inorder(root: Node):
    global cur, mx, node, previous
 
    # If root is null then return
    if root is None:
        return
 
    inorder(root.left)
 
    if previous is not None:
 
        # If the previous value is equal to
        # the current value then increase the count
        if root.val == previous.val:
            cur += 1
 
        # Else initialize the count by 1
        else:
            cur = 1
 
    # If current count is greater than the
    # max count then update the mx value
    if cur > mx:
        mx = cur
        node = root.val
 
    # Make the current Node as previous
    previous = root
    inorder(root.right)
 
# Utility function
def findNode(root: Node) -> int:
    global node
 
    inorder(root)
    return node
 
# Driver Code
if __name__ == "__main__":
    # Let us create following BST
    #         6
    #         / \
    #     5     7
    #     / \ / \
    #     4 5 7 7
    root = newNode(6)
    root.left = newNode(5)
    root.right = newNode(7)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.right.left = newNode(7)
    root.right.right = newNode(7)
 
    print("Node of BST is", findNode(root))
 
# This code is contributed by
# sanjeev2552

C#

/* C# program to find the median of BST 
in O(n) time and O(1) space*/
using System;
     
class GFG
{
     
/* A binary search tree Node has data, pointer
to left child and a pointer to right child */
public class Node
{
    public int val;
    public Node left, right;
};
 
static Node newNode(int data)
{
    Node temp = new Node();
    temp.val = data;
    temp.left = temp.right = null;
    return temp;
}
 
// cur for storing the current count 
// of the value and mx for the maximum count
// of the element which is denoted by node
static int cur = 1, mx = 0;
static int node;
static Node previous = null;
 
// Find the inorder traversal of the BST
static void inorder(Node root)
{
    // If root is null then return
    if (root == null) 
    {
        return;
    }
    inorder(root.left);
    if (previous != null) 
    {
         
        // If the previous value is equal to 
        // the current value then increase the count
        if (root.val == previous.val) 
        {
            cur++;
        }
         
        // Else initialize the count by 1
        else
        {
            cur = 1;
        }
    }
     
    // If current count is greater than the 
    // max count then update the mx value
    if (cur > mx) 
    {
        mx = cur;
        node = root.val;
    }
     
    // Make the current Node as previous
    previous = root;
    inorder(root.right);
}
 
// Utility function
static int findnode(Node root)
{
    inorder(root);
    return node;
}
 
// Driver Code
public static void Main(String []args)
{
    /* Let us create following BST
                6
                / \
                5     7
            / \ / \
            4     5 7 7 
    */
    Node root = newNode(6);
    root.left = newNode(5);
    root.right = newNode(7);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(7);
    root.right.right = newNode(7);
 
    Console.WriteLine("Node of BST is " + 
                         findnode(root));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
 
/* Javascript program to find the median of BST 
in O(n) time and O(1) space*/
     
/* A binary search tree Node has data, pointer
to left child and a pointer to right child */
class Node
{
    constructor()
    {
        this.val = 0;
        this.left = null;
        this.right = null;
    }
};
 
function newNode(data)
{
    var temp = new Node();
    temp.val = data;
    temp.left = temp.right = null;
    return temp;
}
 
// cur for storing the current count 
// of the value and mx for the maximum count
// of the element which is denoted by node
var cur = 1, mx = 0;
var node;
var previous = null;
 
// Find the inorder traversal of the BST
function inorder(root)
{
    // If root is null then return
    if (root == null) 
    {
        return;
    }
    inorder(root.left);
    if (previous != null) 
    {
         
        // If the previous value is equal to 
        // the current value then increase the count
        if (root.val == previous.val) 
        {
            cur++;
        }
         
        // Else initialize the count by 1
        else
        {
            cur = 1;
        }
    }
     
    // If current count is greater than the 
    // max count then update the mx value
    if (cur > mx) 
    {
        mx = cur;
        node = root.val;
    }
     
    // Make the current Node as previous
    previous = root;
    inorder(root.right);
}
 
// Utility function
function findnode(root)
{
    inorder(root);
    return node;
}
 
// Driver Code
/* Let us create following BST
            6
            / \
            5     7
        / \ / \
        4     5 7 7 
*/
var root = newNode(6);
root.left = newNode(5);
root.right = newNode(7);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.left = newNode(7);
root.right.right = newNode(7);
document.write("Node of BST is " + 
                     findnode(root));
 
// This code is contributed by noob2000.
</script>

Output:

node of BST is 7

Time complexity : $O(N)$
Auxiliary Space: O(N).

Count the number of Nodes in a Binary Tree in Constant Space

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Find maximum count of duplicate nodes in a Binary Search Tree

C++

Java

Python3

C#

Javascript

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