draw binary search tree java

Binary Search Tree(BST)

In this tutorial, you will learn how Binary Search Tree works. Also, you will find working examples of Binary Search Tree in C, C++, Java and Python.

Binary search tree is a data structure that quickly allows us to maintain a sorted list of numbers.

  • It is called a binary tree because each tree node has a maximum of two children.
  • It is called a search tree because it can be used to search for the presence of a number in O(log(n)) time.

The properties that separate a binary search tree from a regular binary tree is

  1. All nodes of left subtree are less than the root node
  2. All nodes of right subtree are more than the root node
  3. Both subtrees of each node are also BSTs i.e. they have the above two properties
A tree having a right subtree with one value smaller than the root is shown to demonstrate that it is not a valid binary search tree
A tree having a right subtree with one value smaller than the root is shown to demonstrate that it is not a valid binary search tree

The binary tree on the right isn't a binary search tree because the right subtree of the node "3" contains a value smaller than it.

There are two basic operations that you can perform on a binary search tree:


The algorithm depends on the property of BST that if each left subtree has values below root and each right subtree has values above the root.

If the value is below the root, we can say for sure that the value is not in the right subtree; we need to only search in the left subtree and if the value is above the root, we can say for sure that the value is not in the left subtree; we need to only search in the right subtree.

Algorithm:

            If root == NULL      return NULL; If number == root->data      return root->data; If number < root->data      return search(root->left) If number > root->data      return search(root->right)          

Let us try to visualize this with a diagram.

4 is not found so, traverse through the left subtree of 8
4 is not found so, traverse through the left subtree of 8
4 is not found so, traverse through the right subtree of 3
4 is not found so, traverse through the right subtree of 3
4 is not found so, traverse through the left subtree of 6
4 is not found so, traverse through the left subtree of 6
4 is found
4 is found

If the value is found, we return the value so that it gets propagated in each recursion step as shown in the image below.

If you might have noticed, we have called return search(struct node*) four times. When we return either the new node or NULL, the value gets returned again and again until search(root) returns the final result.

if the value is found in any of the subtrees, it is propagated up so that in the end it is returned, otherwise null is returned
If the value is found in any of the subtrees, it is propagated up so that in the end it is returned, otherwise null is returned

If the value is not found, we eventually reach the left or right child of a leaf node which is NULL and it gets propagated and returned.


Insert Operation

Inserting a value in the correct position is similar to searching because we try to maintain the rule that the left subtree is lesser than root and the right subtree is larger than root.

We keep going to either right subtree or left subtree depending on the value and when we reach a point left or right subtree is null, we put the new node there.

Algorithm:

            If node == NULL      return createNode(data) if (data < node->data)     node->left  = insert(node->left, data); else if (data > node->data)     node->right = insert(node->right, data);   return node;          

The algorithm isn't as simple as it looks. Let's try to visualize how we add a number to an existing BST.

4<8 so, transverse through the left child of 8
4<8 so, transverse through the left child of 8
4>3 so, transverse through the right child of 4
4>3 so, transverse through the right child of 8
4<6 so, transverse through the left child of 6
4<6 so, transverse through the left child of 6
Insert 4 as a left child of 6
Insert 4 as a left child of 6

We have attached the node but we still have to exit from the function without doing any damage to the rest of the tree. This is where the return node; at the end comes in handy. In the case of NULL, the newly created node is returned and attached to the parent node, otherwise the same node is returned without any change as we go up until we return to the root.

This makes sure that as we move back up the tree, the other node connections aren't changed.

Image showing the importance of returning the root element at the end so that the elements don't lose their position during the upward recursion step.
Image showing the importance of returning the root element at the end so that the elements don't lose their position during the upward recursion step.

Deletion Operation

There are three cases for deleting a node from a binary search tree.

Case I

In the first case, the node to be deleted is the leaf node. In such a case, simply delete the node from the tree.

4 is to be deleted
4 is to be deleted
Delete the node
Delete the node

Case II

In the second case, the node to be deleted lies has a single child node. In such a case follow the steps below:

  1. Replace that node with its child node.
  2. Remove the child node from its original position.
6 is to be deleted
6 is to be deleted
copy the value of its child to the node
copy the value of its child to the node and delete the child
Final tree
Final tree

Case III

In the third case, the node to be deleted has two children. In such a case follow the steps below:

  1. Get the inorder successor of that node.
  2. Replace the node with the inorder successor.
  3. Remove the inorder successor from its original position.
3 is to be deleted
3 is to be deleted
Copy the value of the inorder successor (4) to the node
Copy the value of the inorder successor (4) to the node
delete the inorder successor
Delete the inorder successor

Python, Java and C/C++ Examples

                # Binary Search Tree operations in Python   # Create a node class Node:     def __init__(self, key):         self.key = key         self.left = None         self.right = None   # Inorder traversal def inorder(root):     if root is not None:         # Traverse left         inorder(root.left)          # Traverse root         print(str(root.key) + "->", end=' ')          # Traverse right         inorder(root.right)   # Insert a node def insert(node, key):      # Return a new node if the tree is empty     if node is None:         return Node(key)      # Traverse to the right place and insert the node     if key < node.key:         node.left = insert(node.left, key)     else:         node.right = insert(node.right, key)      return node   # Find the inorder successor def minValueNode(node):     current = node      # Find the leftmost leaf     while(current.left is not None):         current = current.left      return current   # Deleting a node def deleteNode(root, key):      # Return if the tree is empty     if root is None:         return root      # Find the node to be deleted     if key < root.key:         root.left = deleteNode(root.left, key)     elif(key > root.key):         root.right = deleteNode(root.right, key)     else:         # If the node is 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          # If the node has two children,         # place the inorder successor in position of the node to be deleted         temp = minValueNode(root.right)          root.key = temp.key          # Delete the inorder successor         root.right = deleteNode(root.right, temp.key)      return root   root = None root = insert(root, 8) root = insert(root, 3) root = insert(root, 1) root = insert(root, 6) root = insert(root, 7) root = insert(root, 10) root = insert(root, 14) root = insert(root, 4)  print("Inorder traversal: ", end=' ') inorder(root)  print("\nDelete 10") root = deleteNode(root, 10) print("Inorder traversal: ", end=' ') inorder(root)              
                // Binary Search Tree operations in Java  class BinarySearchTree {   class Node {     int key;     Node left, right;      public Node(int item) {       key = item;       left = right = null;     }   }    Node root;    BinarySearchTree() {     root = null;   }    void insert(int key) {     root = insertKey(root, key);   }    // Insert key in the tree   Node insertKey(Node root, int key) {     // Return a new node if the tree is empty     if (root == null) {       root = new Node(key);       return root;     }      // Traverse to the right place and insert the node     if (key < root.key)       root.left = insertKey(root.left, key);     else if (key > root.key)       root.right = insertKey(root.right, key);      return root;   }    void inorder() {     inorderRec(root);   }    // Inorder Traversal   void inorderRec(Node root) {     if (root != null) {       inorderRec(root.left);       System.out.print(root.key + " -> ");       inorderRec(root.right);     }   }    void deleteKey(int key) {     root = deleteRec(root, key);   }    Node deleteRec(Node root, int key) {     // Return if the tree is empty     if (root == null)       return root;      // Find the node to be deleted     if (key < root.key)       root.left = deleteRec(root.left, key);     else if (key > root.key)       root.right = deleteRec(root.right, key);     else {       // If the node is with only one child or no child       if (root.left == null)         return root.right;       else if (root.right == null)         return root.left;        // If the node has two children       // Place the inorder successor in position of the node to be deleted       root.key = minValue(root.right);        // Delete the inorder successor       root.right = deleteRec(root.right, root.key);     }      return root;   }    // Find the inorder successor   int minValue(Node root) {     int minv = root.key;     while (root.left != null) {       minv = root.left.key;       root = root.left;     }     return minv;   }    // Driver Program to test above functions   public static void main(String[] args) {     BinarySearchTree tree = new BinarySearchTree();      tree.insert(8);     tree.insert(3);     tree.insert(1);     tree.insert(6);     tree.insert(7);     tree.insert(10);     tree.insert(14);     tree.insert(4);      System.out.print("Inorder traversal: ");     tree.inorder();      System.out.println("\n\nAfter deleting 10");     tree.deleteKey(10);     System.out.print("Inorder traversal: ");     tree.inorder();   } }              
                // Binary Search Tree operations in C  #include <stdio.h> #include <stdlib.h>  struct node {   int key;   struct node *left, *right; };  // Create a 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; }  // Inorder Traversal void inorder(struct node *root) {   if (root != NULL) {     // Traverse left     inorder(root->left);      // Traverse root     printf("%d -> ", root->key);      // Traverse right     inorder(root->right);   } }  // Insert a node struct node *insert(struct node *node, int key) {   // Return a new node if the tree is empty   if (node == NULL) return newNode(key);    // Traverse to the right place and insert the node   if (key < node->key)     node->left = insert(node->left, key);   else     node->right = insert(node->right, key);    return node; }  // Find the inorder successor struct node *minValueNode(struct node *node) {   struct node *current = node;    // Find the leftmost leaf   while (current && current->left != NULL)     current = current->left;    return current; }  // Deleting a node struct node *deleteNode(struct node *root, int key) {   // Return if the tree is empty   if (root == NULL) return root;    // Find the node to be deleted   if (key < root->key)     root->left = deleteNode(root->left, key);   else if (key > root->key)     root->right = deleteNode(root->right, key);    else {     // If the node is 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;     }      // If the node has two children     struct node *temp = minValueNode(root->right);      // Place the inorder successor in position of the node to be deleted     root->key = temp->key;      // Delete the inorder successor     root->right = deleteNode(root->right, temp->key);   }   return root; }  // Driver code int main() {   struct node *root = NULL;   root = insert(root, 8);   root = insert(root, 3);   root = insert(root, 1);   root = insert(root, 6);   root = insert(root, 7);   root = insert(root, 10);   root = insert(root, 14);   root = insert(root, 4);    printf("Inorder traversal: ");   inorder(root);    printf("\nAfter deleting 10\n");   root = deleteNode(root, 10);   printf("Inorder traversal: ");   inorder(root); }              
                // Binary Search Tree operations in C++  #include <iostream> using namespace std;  struct node {   int key;   struct node *left, *right; };  // Create a 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; }  // Inorder Traversal void inorder(struct node *root) {   if (root != NULL) {     // Traverse left     inorder(root->left);      // Traverse root     cout << root->key << " -> ";      // Traverse right     inorder(root->right);   } }  // Insert a node struct node *insert(struct node *node, int key) {   // Return a new node if the tree is empty   if (node == NULL) return newNode(key);    // Traverse to the right place and insert the node   if (key < node->key)     node->left = insert(node->left, key);   else     node->right = insert(node->right, key);    return node; }  // Find the inorder successor struct node *minValueNode(struct node *node) {   struct node *current = node;    // Find the leftmost leaf   while (current && current->left != NULL)     current = current->left;    return current; }  // Deleting a node struct node *deleteNode(struct node *root, int key) {   // Return if the tree is empty   if (root == NULL) return root;    // Find the node to be deleted   if (key < root->key)     root->left = deleteNode(root->left, key);   else if (key > root->key)     root->right = deleteNode(root->right, key);   else {     // If the node is 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;     }      // If the node has two children     struct node *temp = minValueNode(root->right);      // Place the inorder successor in position of the node to be deleted     root->key = temp->key;      // Delete the inorder successor     root->right = deleteNode(root->right, temp->key);   }   return root; }  // Driver code int main() {   struct node *root = NULL;   root = insert(root, 8);   root = insert(root, 3);   root = insert(root, 1);   root = insert(root, 6);   root = insert(root, 7);   root = insert(root, 10);   root = insert(root, 14);   root = insert(root, 4);    cout << "Inorder traversal: ";   inorder(root);    cout << "\nAfter deleting 10\n";   root = deleteNode(root, 10);   cout << "Inorder traversal: ";   inorder(root); }              

Binary Search Tree Complexities

Time Complexity

Operation Best Case Complexity Average Case Complexity Worst Case Complexity
Search O(log n) O(log n) O(n)
Insertion O(log n) O(log n) O(n)
Deletion O(log n) O(log n) O(n)

Here, n is the number of nodes in the tree.

Space Complexity

The space complexity for all the operations is O(n).


Binary Search Tree Applications

  1. In multilevel indexing in the database
  2. For dynamic sorting
  3. For managing virtual memory areas in Unix kernel

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Source: https://www.programiz.com/dsa/binary-search-tree

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