We have discussed BST search and insert operations. In this post, delete operation is discussed. When we delete a node, three possibilities arise.
1) Node to be deleted is leaf: Simply remove from the tree.
50 50
/ \ delete(20) / \
30 70 ---------> 30 70
/ \ / \ \ / \
20 40 60 80 40 60 80
2) Node to be deleted has only one child: Copy the child to the node and delete the child
50 50
/ \ delete(30) / \
30 70 ---------> 40 70
\ / \ / \
40 60 80 60 80
3) Node to be deleted has two children: Find inorder successor of the node. Copy contents of the inorder successor to the node and delete the inorder successor. Note that inorder predecessor can also be used.
50 60
/ \ delete(50) / \
40 70 ---------> 40 70
/ \ \
60 80 80
The important thing to note is, inorder successor is needed only when right child is not empty. In this particular case, inorder successor can be obtained by finding the minimum value in right child of the node.
C/C++
// C program to demonstrate delete operation in binary search tree
#include<stdio.h>
#include<stdlib.h>
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);
printf("%d ", 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->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 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;
// Delete the inorder successor
root->right = deleteNode(root->right, temp->key);
}
return root;
}
// Driver Program to test above functions
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);
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 30\n");
root = deleteNode(root, 30);
printf("Inorder traversal of the modified tree \n");
inorder(root);
printf("\nDelete 50\n");
root = deleteNode(root, 50);
printf("Inorder traversal of the modified tree \n");
inorder(root);
return 0;
}
Java
// Java program to demonstrate delete operation in binary search tree
class BinarySearchTree
{
/* Class containing left and right child of current node and key value*/
class Node
{
int key;
Node left, right;
public Node(int item)
{
key = item;
left = right = null;
}
}
// Root of BST
Node root;
// Constructor
BinarySearchTree()
{
root = null;
}
// This method mainly calls deleteRec()
void deleteKey(int key)
{
root = deleteRec(root, key);
}
/* A recursive function to insert a new key in BST */
Node deleteRec(Node root, int key)
{
/* Base Case: If the tree is empty */
if (root == null) return root;
/* Otherwise, recur down the tree */
if (key < root.key)
root.left = deleteRec(root.left, key);
else if (key > root.key)
root.right = deleteRec(root.right, key);
// if key is same as root's key, then This is the node
// to be deleted
else
{
// node with only one child or no child
if (root.left == null)
return root.right;
else if (root.right == null)
return root.left;
// node with two children: Get the inorder successor (smallest
// in the right subtree)
root.key = minValue(root.right);
// Delete the inorder successor
root.right = deleteRec(root.right, root.key);
}
return root;
}
int minValue(Node root)
{
int minv = root.key;
while (root.left != null)
{
minv = root.left.key;
root = root.left;
}
return minv;
}
// This method mainly calls insertRec()
void insert(int key)
{
root = insertRec(root, key);
}
/* A recursive function to insert a new key in BST */
Node insertRec(Node root, int key)
{
/* If the tree is empty, return a new node */
if (root == null)
{
root = new Node(key);
return root;
}
/* Otherwise, recur down the tree */
if (key < root.key)
root.left = insertRec(root.left, key);
else if (key > root.key)
root.right = insertRec(root.right, key);
/* return the (unchanged) node pointer */
return root;
}
// This method mainly calls InorderRec()
void inorder()
{
inorderRec(root);
}
// A utility function to do inorder traversal of BST
void inorderRec(Node root)
{
if (root != null)
{
inorderRec(root.left);
System.out.print(root.key + " ");
inorderRec(root.right);
}
}
// Driver Program to test above functions
public static void main(String[] args)
{
BinarySearchTree tree = new BinarySearchTree();
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
tree.insert(50);
tree.insert(30);
tree.insert(20);
tree.insert(40);
tree.insert(70);
tree.insert(60);
tree.insert(80);
System.out.println("Inorder traversal of the given tree");
tree.inorder();
System.out.println("\nDelete 20");
tree.deleteKey(20);
System.out.println("Inorder traversal of the modified tree");
tree.inorder();
System.out.println("\nDelete 30");
tree.deleteKey(30);
System.out.println("Inorder traversal of the modified tree");
tree.inorder();
System.out.println("\nDelete 50");
tree.deleteKey(50);
System.out.println("Inorder traversal of the modified tree");
tree.inorder();
}
}
Python
# Python program to demonstrate delete operation
# in binary search tree
# A Binary Tree Node
class Node:
# Constructor to create a new node
def __init__(self, key):
self.key = key
self.left = None
self.right = None
# A utility function to do inorder traversal of BST
def inorder(root):
if root is not None:
inorder(root.left)
print root.key,
inorder(root.right)
# A utility function to insert a new node with given key in BST
def insert( node, key):
# If the tree is empty, return a new node
if node is None:
return Node(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 minum 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 is not None):
current = current.left
return current
# Given a binary search tree and a key, this function
# delete the key and returns the new root
def deleteNode(root, key):
# Base Case
if root is 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 kye to be delete 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, then this is the node
# to be deleted
else:
# Node with only one child or no child
if root.left is None :
temp = root.right
root = None
return temp
elif root.right is None :
temp = root.left
root = None
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
# Delete the inorder successor
root.right = deleteNode(root.right , temp.key)
return root
# Driver program to test above functions
""" Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 """
root = None
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)
print "Inorder traversal of the given tree"
inorder(root)
print "\nDelete 20"
root = deleteNode(root, 20)
print "Inorder traversal of the modified tree"
inorder(root)
print "\nDelete 30"
root = deleteNode(root, 30)
print "Inorder traversal of the modified tree"
inorder(root)
print "\nDelete 50"
root = deleteNode(root, 50)
print "Inorder traversal of the modified tree"
inorder(root)
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
Output:
Inorder traversal of the given tree 20 30 40 50 60 70 80 Delete 20 Inorder traversal of the modified tree 30 40 50 60 70 80 Delete 30 Inorder traversal of the modified tree 40 50 60 70 80 Delete 50 Inorder traversal of the modified tree 40 60 70 80
Illustration:
Time Complexity: The worst case time complexity of delete operation is O(h) where h is height of Binary Search Tree. In worst case, we may have to travel from root to the deepest leaf node. The height of a skewed tree may become n and the time complexity of delete operation may become O(n)
Related Links:
- Binary Search Tree Introduction, Search and Insert/a>
- Quiz on Binary Search Tree
- Coding practice on BST
- All Articles on BST
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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