How to handle duplicates in Binary Search Tree?

• Difficulty Level : Medium
• Last Updated : 14 Dec, 2021

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.
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 on right side (we could also choose left side). For example consider insertion of keys 12, 10, 20, 9, 11, 10, 12, 12 in an empty Binary Search Tree

12
/     \
10      20
/  \     /
9   11   12
/      \
10       12

A Better Solution is to augment every tree node to store count together with regular fields like key, left and right pointers.
Insertion of keys 12, 10, 20, 9, 11, 10, 12, 12 in an empty Binary Search Tree would create following.

12(3)
/        \
10(2)      20(1)
/    \
9(1)   11(1)

Count of a key is shown in bracket

This approach has following advantages over above simple approach.
1) Height of tree is small irrespective of number of duplicates. Note that most of the BST operations (search, insert and delete) have time complexity as O(h) where h is height of BST. So if we are able to keep the height small, we get advantage of less number of key comparisons.
2) Search, Insert and Delete become easier to do. We can use same insert, search and delete algorithms with small modifications (see below code).
3) This approach is suited for self-balancing BSTs (AVL Tree, Red-Black Tree, etc) also. These trees involve rotations, and a rotation may violate BST property of simple solution as a same key can be in either left side or right side after rotation.
Below is implementation of normal Binary Search Tree with count with every key. This code basically is taken from code for insert and delete in BST. The changes made for handling duplicates are highlighted, rest of the code is same.

C++

 // C++ program to implement basic operations// (search, insert and delete) on a BST that// handles duplicates by storing count with// every node#includeusing namespace std; struct node{    int key;    int count;    struct node *left, *right;}; // A utility function to create a new BST nodestruct node *newNode(int item){    struct node *temp = (struct node *)malloc(sizeof(struct node));    temp->key = item;    temp->left = temp->right = NULL;    temp->count = 1;    return temp;} // A utility function to do inorder traversal of BSTvoid inorder(struct node *root){    if (root != NULL)    {        inorder(root->left);        cout << root->key << "("             << root->count << ") ";        inorder(root->right);    }} /* A utility function to insert a newnode with given key in BST */struct node* insert(struct node* node, int key){    /* If the tree is empty, return a new node */    if (node == NULL) return newNode(key);     // If key already exists in BST,    // increment count and return    if (key == node->key)    {    (node->count)++;        return node;    }     /* Otherwise, recur down the tree */    if (key < node->key)        node->left = insert(node->left, key);    else        node->right = insert(node->right, key);     /* return the (unchanged) node pointer */    return node;} /* Given a non-empty binary search tree, returnthe node with minimum key value found in thattree. Note that the entire tree does not needto be searched. */struct node * minValueNode(struct node* node){    struct node* current = node;     /* loop down to find the leftmost leaf */    while (current->left != NULL)        current = current->left;     return current;} /* Given a binary search tree and a key,this function deletes a given key andreturns root of modified tree */struct node* deleteNode(struct node* root,                                 int key){    // base case    if (root == NULL) return root;     // If the key to be deleted is smaller than the    // root's key, then it lies in left subtree    if (key < root->key)        root->left = deleteNode(root->left, key);     // If the key to be deleted is greater than    // the root's key, then it lies in right subtree    else if (key > root->key)        root->right = deleteNode(root->right, key);     // if key is same as root's key    else    {        // If key is present more than once,        // simply decrement count and return        if (root->count > 1)        {            (root->count)--;            return root;        }         // ElSE, delete the node         // node with only one child or no child        if (root->left == NULL)        {            struct node *temp = root->right;            free(root);            return temp;        }        else if (root->right == NULL)        {            struct node *temp = root->left;            free(root);            return temp;        }         // node with two children: Get the inorder        // successor (smallest in the right subtree)        struct node* temp = minValueNode(root->right);         // Copy the inorder successor's        // content to this node        root->key = temp->key;        root->count = temp->count;                 // To ensure successor gets deleted by          // deleteNode call, set count to 0.          temp->count = 0;         // Delete the inorder successor        root->right = deleteNode(root->right,                                  temp->key);    }    return root;} // Driver Codeint main(){    /* Let us create following BST            12(3)        /     \    10(2)     20(1)    / \    9(1) 11(1) */    struct node *root = NULL;    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 20);    root = insert(root, 9);    root = insert(root, 11);    root = insert(root, 10);    root = insert(root, 12);    root = insert(root, 12);     cout << "Inorder traversal of the given tree "         << endl;    inorder(root);     cout << "\nDelete 20\n";    root = deleteNode(root, 20);    cout << "Inorder traversal of the modified tree \n";    inorder(root);     cout << "\nDelete 12\n" ;    root = deleteNode(root, 12);    cout << "Inorder traversal of the modified tree \n";    inorder(root);     cout << "\nDelete 9\n";    root = deleteNode(root, 9);    cout << "Inorder traversal of the modified tree \n";    inorder(root);     return 0;} // This code is contributed by Akanksha Rai

C

 // C program to implement basic operations (search, insert and delete)// on a BST that handles duplicates by storing count with every node#include#include struct node{    int key;    int count;    struct node *left, *right;}; // A utility function to create a new BST nodestruct node *newNode(int item){    struct node *temp =  (struct node *)malloc(sizeof(struct node));    temp->key = item;    temp->left = temp->right = NULL;    temp->count = 1;    return temp;} // A utility function to do inorder traversal of BSTvoid inorder(struct node *root){    if (root != NULL)    {        inorder(root->left);        printf("%d(%d) ", root->key, root->count);        inorder(root->right);    }} /* A utility function to insert a new node with given key in BST */struct node* insert(struct node* node, int key){    /* If the tree is empty, return a new node */    if (node == NULL) return newNode(key);     // If key already exists in BST, increment count and return    if (key == node->key)    {       (node->count)++;        return node;    }     /* Otherwise, recur down the tree */    if (key < node->key)        node->left  = insert(node->left, key);    else        node->right = insert(node->right, key);     /* return the (unchanged) node pointer */    return node;} /* Given a non-empty binary search tree, return the node with   minimum key value found in that tree. Note that the entire   tree does not need to be searched. */struct node * minValueNode(struct node* node){    struct node* current = node;     /* loop down to find the leftmost leaf */    while (current->left != NULL)        current = current->left;     return current;} /* Given a binary search tree and a key, this function   deletes a given key and returns root of modified tree */struct node* deleteNode(struct node* root, int key){    // base case    if (root == NULL) return root;     // If the key to be deleted is smaller than the    // root's key, then it lies in left subtree    if (key < root->key)        root->left = deleteNode(root->left, key);     // If the key to be deleted is greater than the root's key,    // then it lies in right subtree    else if (key > root->key)        root->right = deleteNode(root->right, key);     // if key is same as root's key    else    {        // If key is present more than once, simply decrement        // count and return        if (root->count > 1)        {           (root->count)--;           return root;        }         // ElSE, delete the node         // node with only one child or no child        if (root->left == NULL)        {            struct node *temp = root->right;            free(root);            return temp;        }        else if (root->right == NULL)        {            struct node *temp = root->left;            free(root);            return temp;        }         // node with two children: Get the inorder successor (smallest        // in the right subtree)        struct node* temp = minValueNode(root->right);         // Copy the inorder successor's content to this node        root->key = temp->key;        root->count = temp->count;         // Delete the inorder successor        root->right = deleteNode(root->right, temp->key);    }    return root;} // Driver Program to test above functionsint main(){    /* Let us create following BST             12(3)          /        \       10(2)      20(1)       /   \    9(1)  11(1)   */    struct node *root = NULL;    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 20);    root = insert(root, 9);    root = insert(root, 11);    root = insert(root, 10);    root = insert(root, 12);    root = insert(root, 12);     printf("Inorder traversal of the given tree \n");    inorder(root);     printf("\nDelete 20\n");    root = deleteNode(root, 20);    printf("Inorder traversal of the modified tree \n");    inorder(root);     printf("\nDelete 12\n");    root = deleteNode(root, 12);    printf("Inorder traversal of the modified tree \n");    inorder(root);     printf("\nDelete 9\n");    root = deleteNode(root, 9);    printf("Inorder traversal of the modified tree \n");    inorder(root);     return 0;}

Java

 // Java program to implement basic operations// (search, insert and delete) on a BST that// handles duplicates by storing count with// every nodeclass GFG{static class node{    int key;    int count;    node left, right;}; // A utility function to create a new BST nodestatic node newNode(int item){    node temp = new node();    temp.key = item;    temp.left = temp.right = null;    temp.count = 1;    return temp;} // A utility function to do inorder traversal of BSTstatic void inorder(node root){    if (root != null)    {        inorder(root.left);        System.out.print(root.key + "(" +                         root.count + ") ");        inorder(root.right);    }} /* A utility function to insert a newnode with given key in BST */static node insert(node node, int key){    /* If the tree is empty, return a new node */    if (node == null) return newNode(key);     // If key already exists in BST,    // increment count and return    if (key == node.key)    {    (node.count)++;        return node;    }     /* Otherwise, recur down the tree */    if (key < node.key)        node.left = insert(node.left, key);    else        node.right = insert(node.right, key);     /* return the (unchanged) node pointer */    return node;} /* Given a non-empty binary search tree, returnthe node with minimum key value found in thattree. Note that the entire tree does not needto be searched. */static node minValueNode(node node){    node current = node;     /* loop down to find the leftmost leaf */    while (current.left != null)        current = current.left;     return current;} /* Given a binary search tree and a key,this function deletes a given key andreturns root of modified tree */static node deleteNode(node root, int key){    // base case    if (root == null) return root;     // If the key to be deleted is smaller than the    // root's key, then it lies in left subtree    if (key < root.key)        root.left = deleteNode(root.left, key);     // If the key to be deleted is greater than    // the root's key, then it lies in right subtree    else if (key > root.key)        root.right = deleteNode(root.right, key);     // if key is same as root's key    else    {        // If key is present more than once,        // simply decrement count and return        if (root.count > 1)        {            (root.count)--;            return root;        }         // ElSE, delete the node         // node with only one child or no child        if (root.left == null)        {            node temp = root.right;            root=null;            return temp;        }        else if (root.right == null)        {            node temp = root.left;            root = null;            return temp;        }         // node with two children: Get the inorder        // successor (smallest in the right subtree)        node temp = minValueNode(root.right);         // Copy the inorder successor's        // content to this node        root.key = temp.key;        root.count = temp.count;         // Delete the inorder successor        root.right = deleteNode(root.right,                                temp.key);    }    return root;} // Driver Codepublic static void main(String[] args){    /* Let us create following BST            12(3)        /     \    10(2)     20(1)    / \    9(1) 11(1) */    node root = null;    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 20);    root = insert(root, 9);    root = insert(root, 11);    root = insert(root, 10);    root = insert(root, 12);    root = insert(root, 12);     System.out.print("Inorder traversal of " +                     "the given tree " + "\n");    inorder(root);     System.out.print("\nDelete 20\n");    root = deleteNode(root, 20);    System.out.print("Inorder traversal of " +                     "the modified tree \n");    inorder(root);     System.out.print("\nDelete 12\n");    root = deleteNode(root, 12);    System.out.print("Inorder traversal of " +                     "the modified tree \n");    inorder(root);     System.out.print("\nDelete 9\n");    root = deleteNode(root, 9);    System.out.print("Inorder traversal of " +                     "the modified tree \n");    inorder(root);}} // This code is contributed by 29AjayKumar

Python3

 # Python3 program to implement basic operations# (search, insert and delete) on a BST that handles# duplicates by storing count with every node # A utility function to create a new BST nodeclass newNode:     # Constructor to create a new node    def __init__(self, data):        self.key = data        self.count = 1        self.left = None        self.right = None # A utility function to do inorder# traversal of BSTdef inorder(root):    if root != None:        inorder(root.left)        print(root.key,"(", root.count,")",                                 end = " ")        inorder(root.right) # A utility function to insert a new node# with given key in BSTdef insert(node, key):         # If the tree is empty, return a new node    if node == None:        k = newNode(key)        return k     # If key already exists in BST, increment    # count and return    if key == node.key:        (node.count) += 1        return node     # Otherwise, recur down the tree    if key < node.key:        node.left = insert(node.left, key)    else:        node.right = insert(node.right, key)     # return the (unchanged) node pointer    return node # Given a non-empty binary search tree, return# the node with minimum key value found in that# tree. Note that the entire tree does not need# to be searched.def minValueNode(node):    current = node     # loop down to find the leftmost leaf    while current.left != None:        current = current.left     return current # Given a binary search tree and a key,# this function deletes a given key and# returns root of modified treedef deleteNode(root, key):         # base case    if root == None:        return root     # If the key to be deleted is smaller than the    # root's key, then it lies in left subtree    if key < root.key:        root.left = deleteNode(root.left, key)     # If the key to be deleted is greater than    # the root's key, then it lies in right subtree    elif key > root.key:        root.right = deleteNode(root.right, key)     # if key is same as root's key    else:                 # If key is present more than once,        # simply decrement count and return        if root.count > 1:            root.count -= 1            return root                 # ElSE, delete the node node with        # only one child or no child        if root.left == None:            temp = root.right            return temp        elif root.right == None:            temp = root.left            return temp         # node with two children: Get the inorder        # successor (smallest in the right subtree)        temp = minValueNode(root.right)         # Copy the inorder successor's content        # to this node        root.key = temp.key        root.count = temp.count         # Delete the inorder successor        root.right = deleteNode(root.right, temp.key)    return root # Driver Codeif __name__ == '__main__':         # Let us create following BST    # 12(3)    # / \    # 10(2) 20(1)    # / \    # 9(1) 11(1)    root = None    root = insert(root, 12)    root = insert(root, 10)    root = insert(root, 20)    root = insert(root, 9)    root = insert(root, 11)    root = insert(root, 10)    root = insert(root, 12)    root = insert(root, 12)     print("Inorder traversal of the given tree")    inorder(root)    print()         print("Delete 20")    root = deleteNode(root, 20)    print("Inorder traversal of the modified tree")    inorder(root)    print()     print("Delete 12")    root = deleteNode(root, 12)    print("Inorder traversal of the modified tree")    inorder(root)    print()     print("Delete 9")    root = deleteNode(root, 9)    print("Inorder traversal of the modified tree")    inorder(root) # This code is contributed by PranchalK

C#

 // C# program to implement basic operations// (search, insert and delete) on a BST that// handles duplicates by storing count with// every nodeusing System; class GFG{public class node{    public int key;    public int count;    public node left, right;}; // A utility function to create// a new BST nodestatic node newNode(int item){    node temp = new node();    temp.key = item;    temp.left = temp.right = null;    temp.count = 1;    return temp;} // A utility function to do inorder// traversal of BSTstatic void inorder(node root){    if (root != null)    {        inorder(root.left);        Console.Write(root.key + "(" +                      root.count + ") ");        inorder(root.right);    }} /* A utility function to insert a newnode with given key in BST */static node insert(node node, int key){    /* If the tree is empty,    return a new node */    if (node == null) return newNode(key);     // If key already exists in BST,    // increment count and return    if (key == node.key)    {        (node.count)++;        return node;    }     /* Otherwise, recur down the tree */    if (key < node.key)        node.left = insert(node.left, key);    else        node.right = insert(node.right, key);     /* return the (unchanged) node pointer */    return node;} /* Given a non-empty binary search tree,return the node with minimum key valuefound in that tree. Note that the entire treedoes not need to be searched. */static node minValueNode(node node){    node current = node;     /* loop down to find the leftmost leaf */    while (current.left != null)        current = current.left;     return current;} /* Given a binary search tree and a key,this function deletes a given key andreturns root of modified tree */static node deleteNode(node root, int key){    // base case    if (root == null) return root;     // If the key to be deleted is smaller than the    // root's key, then it lies in left subtree    if (key < root.key)        root.left = deleteNode(root.left, key);     // If the key to be deleted is greater than    // the root's key, then it lies in right subtree    else if (key > root.key)        root.right = deleteNode(root.right, key);     // if key is same as root's key    else    {        // If key is present more than once,        // simply decrement count and return        if (root.count > 1)        {            (root.count)--;            return root;        }         // ElSE, delete the node        node temp = null;                 // node with only one child or no child        if (root.left == null)        {            temp = root.right;            root = null;            return temp;        }        else if (root.right == null)        {            temp = root.left;            root = null;            return temp;        }         // node with two children: Get the inorder        // successor (smallest in the right subtree)        temp = minValueNode(root.right);         // Copy the inorder successor's        // content to this node        root.key = temp.key;        root.count = temp.count;         // Delete the inorder successor        root.right = deleteNode(root.right,                                temp.key);    }    return root;} // Driver Codepublic static void Main(String[] args){    /* Let us create following BST            12(3)        /     \    10(2)     20(1)    / \    9(1) 11(1) */    node root = null;    root = insert(root, 12);    root = insert(root, 10);    root = insert(root, 20);    root = insert(root, 9);    root = insert(root, 11);    root = insert(root, 10);    root = insert(root, 12);    root = insert(root, 12);     Console.Write("Inorder traversal of " +                  "the given tree " + "\n");    inorder(root);     Console.Write("\nDelete 20\n");    root = deleteNode(root, 20);    Console.Write("Inorder traversal of " +                  "the modified tree \n");    inorder(root);     Console.Write("\nDelete 12\n");    root = deleteNode(root, 12);    Console.Write("Inorder traversal of " +                  "the modified tree \n");    inorder(root);     Console.Write("\nDelete 9\n");    root = deleteNode(root, 9);    Console.Write("Inorder traversal of " +                  "the modified tree \n");    inorder(root);}} // This code is contributed by Rajput-Ji

Javascript



Output:

Inorder traversal of the given tree
9(1) 10(2) 11(1) 12(3) 20(1)
Delete 20
Inorder traversal of the modified tree
9(1) 10(2) 11(1) 12(3)
Delete 12
Inorder traversal of the modified tree
9(1) 10(2) 11(1) 12(2)
Delete 9
Inorder traversal of the modified tree
10(2) 11(1) 12(2)

We will soon be discussing AVL and Red Black Trees with duplicates allowed.