remove in bst

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struct Node{
    Node *left, *right;
    int val;
    Node(int data){
        val=data;
        left=right=NULL;
    }
};

Node* inordersuccessor(Node* root){
    Node* curr=root;
    while(curr!=NULL && curr->left!=NULL){
        curr=curr->left;
    }
    return curr;
}

Node* remove(Node* root, int x){
    if(root==NULL) return root;
    else if(x<root->val){
        root->left=remove(root->left,x);
    }
    else if(x>root->val){
        root->right=remove(root->right,x);
    }
    else{
        if(root->left==NULL){
            Node* temp=root->right;
            delete(root);
            return temp;
        }
        else if(root->right==NULL){
            Node* temp=root->left;
            delete(root);
            return temp;
        }
        else{
            Node* ins=inordersuccessor(root->right);
            root->val=ins->val;
            root->right=remove(root->right,ins->val);
        }
        
    }
    return root;
}

Popularity 9/10 Helpfulness 1/10 Language cpp

Source: Grepper

Tags: c++

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Contributed on Oct 29 2022

Homely Hawk

0 Answers Avg Quality 2/10

Deletion in binary search tree (bst)

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/*
 * Program  : Binary Search Tree Deletion
 * Language : C
 */

#include<stdio.h>
#include<stdlib.h>

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

struct node *getNewNode(int val)
{
    struct node *newNode = malloc(sizeof(struct node));
    newNode->key   = val;
    newNode->left  = NULL;
    newNode->right = NULL;

    return newNode;
}

struct node *insert(struct node *root, int val)
{
    if(root == NULL)
        return getNewNode(val);
    if(root->key < val)
        root->right = insert(root->right,val);
    else if(root->key > val)
        root->left = insert(root->left,val);

    return root;
}

int getRightMin(struct node *root)
{
    struct node *temp = root;

    //min value should be present in the left most node.
    while(temp->left != NULL){ temp = temp->left;}

    return temp->key;
}

struct node *removeNode(struct node *root, int val)
{
    /*
     * If the node becomes NULL, it will return NULL
     * Two possible ways which can trigger this case
     * 1. If we send the empty tree. i.e root == NULL
     * 2. If the given node is not present in the tree.
     */
    if(root == NULL)
        return NULL;
    /*
     * If root->key < val. val must be present in the right subtree
     * So, call the above remove function with root->right
     */
    if(root->key < val)
        root->right = removeNode(root->right,val);
    /*
     * if root->key > val. val must be present in the left subtree
     * So, call the above function with root->left
     */
    else if(root->key > val)
        root->left = removeNode(root->left,val);
    /*
     * This part will be executed only if the root->key == val
     * The actual removal starts from here
     */
    else
    {
        /*
         * Case 1: Leaf node. Both left and right reference is NULL
         * replace the node with NULL by returning NULL to the calling pointer.
         * free the node
         */
        if(root->left == NULL && root->right == NULL)
        {
            free(root);
            return NULL;
        }
        /*
         * Case 2: Node has right child.
         * replace the root node with root->right and free the right node
         */
        else if(root->left == NULL)
        {
            struct node *temp = root->right;
            free(root);
            return temp;
        }
        /*
         * Case 3: Node has left child.
         * replace the node with root->left and free the left node
         */
        else if(root->right == NULL)
        {
            struct node *temp = root->left;
            free(root);
            return temp;
        }
        /*
         * Case 4: Node has both left and right children.
         * Find the min value in the right subtree
         * replace node value with min.
         * And again call the remove function to delete the node which has the min value.
         * Since we find the min value from the right subtree call the remove function with root->right.
         */
        else
        {
            int rightMin = getRightMin(root->right);
            root->key = rightMin;
            root->right = removeNode(root->right,rightMin);
        }

    }

    //return the actual root's address
    return root;
}

/*
 * it will print the tree in ascending order
 */
void inorder(struct node *root)
{
    if(root == NULL)
        return;
    inorder(root->left);
    printf("%d ",root->key);
    inorder(root->right);
}

int main()
{
    /*
         100
         / \
        50  200
            /  \
           150  300
     */
    struct node *root = NULL;
    root = insert(root,100);
    root = insert(root,50);
    root = insert(root,200);
    root = insert(root,150);
    root = insert(root,300);

    printf("Initial tree :\t");
    inorder(root);
    printf("\n");

    /* remove leaf node 300
         100
         / \
        50  200
            /
           150
     */
    root = removeNode(root,300);
    printf("After deletion of 300, the new tree :\t");
    inorder(root);
    printf("\n");

    /* remove root node 100
         150
         / \
        50  200
     */
    root = removeNode(root,100);
    printf("After deletion of 100, the new tree :\t");
    inorder(root);
    printf("\n");

    return 0;
}

Popularity 9/10 Helpfulness 2/10 Language whatever

Source: www.log2base2.com

Tags: binary-search-tree binary-search-tre

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Contributed on Dec 02 2022

beingarifkhan

0 Answers Avg Quality 2/10

Deletion in binary search tree (bst)

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(step 2)           (step 3)
  100                                           100                 100
  / \                                           / \                 / \
50  200         delete(200)                    50  300             50  300
    / \         (right subtree min = 300)          / \                 /
  150  300      (step 1)                          150 300             150

Popularity 9/10 Helpfulness 1/10 Language whatever

Source: www.log2base2.com

Tags: binary-search-tree binary-search-tre

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Contributed on Dec 02 2022

beingarifkhan

0 Answers Avg Quality 2/10

deletion in bst

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// C++ program to demonstrate 
// delete operation in binary
// search tree
#include <bits/stdc++.h>
using namespace std;

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

// A utility function to create a new BST node
struct node* newNode(int item)
{
    struct node* temp
        = (struct node*)malloc(sizeof(struct node));
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}

// A utility function to do 
// inorder traversal of BST
void inorder(struct node* root)
{
    if (root != NULL) {
        inorder(root->left);
        cout << root->key;
        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);

    /* 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 && current->left != NULL)
        current = current->left;

    return current;
}

/* Given a binary search tree and a key, this function
deletes the key and returns the new root */
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, then This is the node
    // to be deleted
    else {
        // node has no child
        if (root->left==NULL and root->right==NULL)
            return NULL;
      
        // node with only one child or no child
        else 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;

        // Delete the inorder successor
        root->right = deleteNode(root->right, temp->key);
    }
    return root;
}

// Driver Code
int main()
{
    /* Let us create following BST
            50
        /     \
        30     70
        / \ / \
    20 40 60 80 */
    struct node* root = NULL;
    root = insert(root, 50);
    root = insert(root, 30);
    root = insert(root, 20);
    root = insert(root, 40);
    root = insert(root, 70);
    root = insert(root, 60);
    root = insert(root, 80);

    cout << "Inorder traversal of the given tree \n";
    inorder(root);

    cout << "\nDelete 20\n";
    root = deleteNode(root, 20);
    cout << "Inorder traversal of the modified tree \n";
    inorder(root);

    cout << "\nDelete 30\n";
    root = deleteNode(root, 30);
    cout << "Inorder traversal of the modified tree \n";
    inorder(root);

    cout << "\nDelete 50\n";
    root = deleteNode(root, 50);
    cout << "Inorder traversal of the modified tree \n";
    inorder(root);

    return 0;
}

// This code is contributed by shivanisinghss2110

Popularity 10/10 Helpfulness 1/10 Language cpp

Source: www.geeksforgeeks.org

Tags: c++

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Contributed on Jan 14 2022

Easy Elk

0 Answers Avg Quality 2/10

remove in bst

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