In a Binary Search Tree (BST), all keys in left subtree of a key must be smaller and all keys in right subtree must be greater. So a Binary Search Tree by definition has distinct keys and duplicates in binary search tree are not allowed.

How to allow duplicates where every insertion inserts one more key with a value and every deletion deletes one occurrence?

A **Simple Solution** is to allow same keys with count. For example consider insertion of keys 3, 6, 7, 8, 8, 8, 10, 12, 12 in an empty Binary Search Tree

8(3) / \ 6(1) 10(1) / \ \ 3(1) 7(1) 12(2)

This count store requires changing the structure of Binary Search tree to

struct node { int key; int count; struct node *left, *right; };

And We may need to change the code of Functions of “Insert” “Find” and “Delete”.

Lets see for Insert

struct node* insert(struct node* node, int key) { if (node == NULL) return newNode(key); if (key == node->key) { (node->count)++; return node; } if (key < node->key) node->left = insert(node->left, key); else node->right = insert(node->right, key); return node; }

Now Based upon Above code You can try to write code for “Search a value in BST with Duplicate” and Delete a node in BST with duplicate”.

PS : In delete a node in BST, Handling of adjusment of tree branches should happen.

**Solution 2 :**

If change in Structure of Binary Search Tree is not allowed, We can think of Extra memory like Hashmap to keep the count of nodes of BST to support duplicates in Binary Search Tree. ie Hashmap below is with above tree without count.

3 - 1 6 - 1 7 - 1 8 - 3 10 - 1 12 - 2

**Solution 3:**

We can think of keeping same keys on left side (we could also choose right side) to support duplicates in Binary Search Tree. For example consider insertion of keys 3, 6, 7, 8, 8, 8, 10, 12, 12 in an empty Binary Search Tree

8 / \ 8 12 / / \ 8 10 12 / 6 / \ 3 7

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