# Find maximum count of duplicate nodes in a Binary Search Tree

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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ program to find the median of BST in O(n)  ` `   ``time and O(1) space*/` ` `  `#include ` `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 currrent 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 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 currrent 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 `

 `/* 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 currrent 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; ` `} ` ` `  `// Drive 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 `

Output:
```node of BST is 7
```

Time complexity :

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : andrew1234, princiraj1992

Article Tags :