Binary Search Tree | Set 3 (Iterative Delete) - GeeksforGeeks

26 Jul, 2021
Given a binary search tree and a node of the binary search tree, the task is to delete the node from the Binary Search tree Iteratiely.Here are the three cases that arise while performing a delete operation on a BST: 
1. Case 1: Node to be deleted is a leaf node. Directly delete the node from the tree. 
         
             10                             10
           /     \         delete(5)      /     \
          7       15       --------->    7       15 
         /  \    /  \                     \     /  \ 
        5    8  11   18                    8   11   18
2. Case 2: Node to be deleted is an internal node with two children. Copy the contents of the inorder successor of the node to be deleted and delete the inorder successor. The inorder successor can be found by finding the minimum element in the right subtree of the node. inorderSuccessor(10) = 11. 
             10                              11
           /     \         delete(10)      /     \
          7       15       --------->    7        15 
         /  \    /  \                   /  \        \ 
        5    8  11   18                5    8        18
3. Case 3: Node to be deleted is an internal node with one child. For this case, delete the node and move its child up to take its place. 
             10                              10
           /     \         delete(15)      /     \
          7       15       --------->    7       11 
         /  \    /                      /  \      
        5    8  11                     5    8
The intuition behind deleting the inorder successor in Case 2 is that the inorder successor of a node with two children will always be greater than all elements in the left sub-tree of the of the node since it is the smallest node in the right sub-tree of the node and the inorder successor of the node will always be smaller than all other nodes in the right sub-tree of the node. 
This preserves the BST property of all nodes in the left sub-tree  of a given node are smaller than the given node and all nodes in the right sub-tree of the given node are greater than the given node. 
Below is the implementation of the above approach: 
C++
Python3
// C++ implementation to delete// a node in the BST #include <bits/stdc++.h>using namespace std; // Structure of the nodetypedef struct treeNode {    int data;    struct treeNode* left;    struct treeNode* right;} treeNode; // Utility function to print// the inorder traversal of the BST.void inorder(treeNode* root){    if (root != NULL) {        inorder(root->left);        cout << root->data << ' ';        inorder(root->right);    }} // Utility function to insert// nodes into our BSTtreeNode* insert(treeNode* root, int key){    // Check if tree is empty    if (root == NULL) {        treeNode* temp;        temp = (treeNode*)malloc(sizeof(treeNode));        temp->data = key;        temp->left = NULL;        temp->right = NULL;        return temp;    }    if (key < root->data) {         // if the key to be inserted        // is lesser than the root,        // insert into the left subtree,        // and recursively call        // the insert function with the        // root->left as the new root.        root->left = insert(root->left, key);    }    else {         // if the key to be inserted        // is greater than the root,        // insert into the right subtree,        // and recursively call        // the insert function with the        // root->right as the new root.        root->right = insert(root->right, key);    }    return root;} // Iterative Function to delete// 'key' from the BST.treeNode* deleteIterative(treeNode* root, int key){    treeNode* curr = root;    treeNode* prev = NULL;     // Check if the key is actually    // present in the BST.    // the variable prev points to    // the parent of the key to be deleted.    while (curr != NULL && curr->data != key) {        prev = curr;        if (key < curr->data)            curr = curr->left;        else            curr = curr->right;    }     if (curr == NULL) {        cout << "Key " << key << " not found in the"             << " provided BST.\n";        return root;    }     // Check if the node to be    // deleted has atmost one child.    if (curr->left == NULL || curr->right == NULL) {         // newCurr will replace        // the node to be deleted.        treeNode* newCurr;         // if the left child does not exist.        if (curr->left == NULL)            newCurr = curr->right;        else            newCurr = curr->left;         // check if the node to        // be deleted is the root.        if (prev == NULL)            return newCurr;         // check if the node to be deleted        // is prev's left or right child        // and then replace this with newCurr        if (curr == prev->left)            prev->left = newCurr;        else            prev->right = newCurr;         // free memory of the        // node to be deleted.        free(curr);    }     // node to be deleted has    // two children.    else {        treeNode* p = NULL;        treeNode* temp;         // Compute the inorder successor        temp = curr->right;        while (temp->left != NULL) {            p = temp;            temp = temp->left;        }         // check if the parent of the inorder        // successor is the curr or not(i.e. curr=        // the node which has the same data as        // the given data by the user to be        // deleted). if it isn't, then make the        // the left child of its parent equal to        // the inorder successor'd right child.        if (p != NULL)            p->left = temp->right;         // if the inorder successor was the        // curr (i.e. curr = the node which has the        // same data as the given data by the        // user to be deleted), then make the        // right child of the node to be        // deleted equal to the right child of        // the inorder successor.        else            curr->right = temp->right;         curr->data = temp->data;        free(temp);    }    return root;} // Driver Codeint main(){    /*         10        /  \       7    15      / \   / \      5  8 11 18     */    treeNode* root = NULL;    root = insert(root, 10);    root = insert(root, 7);    root = insert(root, 5);    root = insert(root, 8);    root = insert(root, 15);    root = insert(root, 11);    root = insert(root, 18);     cout << "Inorder traversal "         << "of original BST:\n";    inorder(root);    cout << '\n';     // delete node with data value 11 (leaf)    root = deleteIterative(root, 11);    cout << "\nDeletion of 11\n";    cout << "Inorder traversal post deletion:\n";    inorder(root);    cout << '\n';     // delete node with data value 15    // (internal node with one child)    root = deleteIterative(root, 15);    cout << "\nDeletion of 15\n";    cout << "Inorder traversal post deletion:\n";    inorder(root);    cout << '\n';     // delete node with data value 10    // (root, two children)    root = deleteIterative(root, 10);    cout << "\nDeletion of 10\n";    cout << "Inorder traversal post deletion:\n";    inorder(root);    cout << '\n';     return 0;}
# Python implementation to delete# a node in the Binary Search Tree # Class for a node of BST.  class Node:    def __init__(self, data):        self.data = data        self.left = None        self.right = None # Utility function to print# the inorder traversal of the BST  def inorder(root):    if root != None:        inorder(root.left)        print(root.data, end=" ")        inorder(root.right) # Utility function to insert# nodes into our BST  def insert(root, key):    # check if tree is empty    if root == None:        temp = Node(key)        return temp     if key < root.data:         """        if the key to be inserted is        lesser than the root,        insert into the left subtree,        and recursively call        the insert function with        the root.left as the new root.        """        root.left = insert(root.left, key)     else:        """        if the key to be inserted is        greater than the root,        insert into the right subtree,        and recursively call        the insert function with the        root->right as the new root.        """        root.right = insert(root.right, key)     return root # Iterative approach to# delete 'key' from the BST.  def deleteIterative(root, key):    curr = root    prev = None     # First check if the key is    # actually present in the BST.    # the variable prev points to the    # parent of the key to be deleted    while(curr != None and curr.data != key):        prev = curr        if curr.data < key:            curr = curr.right        else:            curr = curr.left     if curr == None:        print("Key % d not found in\           the provided BST." % key)        return root     # Check if the node to be    # deleted has atmost one child    if curr.left == None or\            curr.right == None:         # newCurr will replace        # the node to be deleted.        newCurr = None         # if the left child does not exist.        if curr.left == None:            newCurr = curr.right        else:            newCurr = curr.left         # check if the node to        # be deleted is the root.        if prev == None:            return newCurr         # Check if the node to be        # deleted is prev's left or        # right child and then        # replace this with newCurr        if curr == prev.left:            prev.left = newCurr        else:            prev.right = newCurr         curr = None     # node to be deleted    # has two children.    else:        p = None        temp = None         # Compute the inorder        # successor of curr.        temp = curr.right        while(temp.left != None):            p = temp            temp = temp.left         # check if the parent of the        # inorder successor is the root or not.        # if it isn't, then make the left        # child of its parent equal to the        # inorder successor's right child.        if p != None:            p.left = temp.right         else:             # if the inorder successor was            # the root, then make the right child            # of the node to be deleted equal            # to the right child of the inorder            # successor.            curr.right = temp.right         curr.data = temp.data        temp = None     return root # Function to create the BST# and call the Delete Function  def main():    """         10                /   \               7     15          / \   / \          5 8  11 18        """    root = None    root = insert(root, 10)    root = insert(root, 7)    root = insert(root, 5)    root = insert(root, 8)    root = insert(root, 15)    root = insert(root, 11)    root = insert(root, 18)     print("Inorder traversal of original BST:")    inorder(root)    print("\n")     # delete node with data value 11 (leaf)    root = deleteIterative(root, 11)    print("Deletion of 11")    print("Inorder traversal post deletion:")    inorder(root)    print("\n")     # delete node with data value 15    # (internal node with one child)    root = deleteIterative(root, 15)    print("Deletion of 15")    print("Inorder traversal post deletion:")    inorder(root)    print("\n")     # delete node with data value 10    # (root, two children)    root = deleteIterative(root, 10)    print("Deletion of 10")    print("Inorder traversal post deletion:")    inorder(root)    print()  # Driver Codeif __name__ == "__main__":    main()
Inorder traversal of original BST:
5 7 8 10 11 15 18 

Deletion of 11
Inorder traversal post deletion:
5 7 8 10 15 18 

Deletion of 15
Inorder traversal post deletion:
5 7 8 10 18 

Deletion of 10
Inorder traversal post deletion:
5 7 8 18
 
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