BST Delete Operation in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the minimum value node in a BST.

DSA Javascript

function findMinNode(node) {
  while (node.[1] !== null) {
    node = node.[1];
  }
  return node;
}

Drag options to blanks, or click blank then click option'

Achild

Bright

Cparent

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' will find the maximum, not minimum.

In a BST, the minimum value is found by going to the left child repeatedly until null.

2fill in blank

medium

Complete the code to delete a node with no children (leaf node) in BST.

DSA Javascript

if (node.left === null && node.right === null) {
  node = [1];
}

Drag options to blanks, or click blank then click option'

Anull

Bnode.right

Cnode.parent

Dnode.left

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning node.left or node.right instead of null.

Deleting a leaf node means setting it to null.

3fill in blank

hard

Fix the error in the code to correctly replace a node with its right child when left child is null.

DSA Javascript

if (node.left === null) {
  node = node.[1];
}

Drag options to blanks, or click blank then click option'

Aright

Bchild

Cleft

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' causes wrong replacement.

If the left child is null, replace the node with its right child.

4fill in blank

hard

Fill both blanks to correctly find the inorder successor and replace the node's value.

DSA Javascript

let temp = findMinNode(node.[1]);
node.[2] = temp.data;

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cdata

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using left subtree or wrong property name for replacement.

The inorder successor is the minimum node in the right subtree. Replace node's data with successor's data.

5fill in blank

hard

Fill all three blanks to complete the recursive delete function for BST.

DSA Javascript

function deleteNode(node, key) {
  if (node === null) return null;
  if (key < node.data) {
    node.[1] = deleteNode(node.[2], key);
  } else if (key > node.data) {
    node.right = deleteNode(node.right, key);
  } else {
    if (node.left === null) return node.[3];
    else if (node.right === null) return node.left;
    let temp = findMinNode(node.right);
    node.data = temp.data;
    node.right = deleteNode(node.right, temp.data);
  }
  return node;
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using right subtree for keys less than node data.

Returning wrong child when left is null.

When key is less, recurse on left subtree. Replace node.left with recursive call result. If left child is null, return right child.

Practice

1. What happens when you delete a leaf node (a node with no children) from a Binary Search Tree (BST)?

easy

A. The node is simply removed without affecting other nodes.

B. The entire tree is rebuilt from scratch.

C. The node's children replace it in the tree.

D. The root node is always changed.

Solution

1. Step 1: Understand leaf node deletion

   A leaf node has no children, so removing it does not affect other nodes.

2. Step 2: Effect on BST structure

   Since no children exist, simply remove the node and update its parent's pointer to null.

3. Final Answer:

   The node is simply removed without affecting other nodes. -> Option A

4. Quick Check:

   Leaf node deletion = simple removal [OK]

Hint: Leaf nodes delete by direct removal only [OK]

Common Mistakes:

• Thinking children replace the node when none exist
• Assuming root always changes
• Believing tree rebuild is needed

2. Which of the following is the correct syntax to delete a node with value key from a BST in JavaScript?

easy

A. bst.deleteNode(key);

B. bst.delete(key);

C. bst.remove(key);

D. bst.erase(key);

Solution

1. Step 1: Identify common BST delete method

   In many JavaScript BST implementations, the method to delete a node is named delete.

2. Step 2: Check method naming conventions

   Methods like remove, erase, or deleteNode are not standard or common in JavaScript BST examples.

3. Final Answer:

   bst.delete(key); -> Option B

4. Quick Check:

   Standard BST delete method = delete() [OK]

Hint: Use bst.delete(key) to remove nodes [OK]

Common Mistakes:

• Using remove() or erase() which are not standard
• Confusing delete keyword with method name
• Using deleteNode() which is uncommon

3. Consider this BST and the delete operation:

    10
   /  \
  5    15
       / \
      12  20

What will be the in-order traversal output after deleting node with value 15?

medium

A. [5, 10, 12, 15, 20]

B. [5, 10, 20]

C. [5, 10, 12, 20]

D. [5, 10, 15, 20]

Solution

1. Step 1: Identify node to delete and its children

   Node 15 has two children: 12 (left) and 20 (right).

2. Step 2: Replace node 15 with its in-order successor

   The in-order successor is the smallest node in right subtree, which is 20. Replace 15 with 20 and delete original 20 node.

3. Step 3: Perform in-order traversal

   In-order traversal visits nodes in ascending order: 5, 10, 12, 20.

4. Final Answer:

   [5, 10, 12, 20] -> Option C

5. Quick Check:

   In-order after deleting 15 = [5, 10, 12, 20] [OK]

Hint: Replace deleted node with in-order successor [OK]

Common Mistakes:

• Removing 15 without replacement
• Using in-order predecessor instead of successor
• Forgetting to delete the successor node

4. The following code attempts to delete a node from a BST but causes an error. What is the main issue?

function deleteNode(root, key) {
  if (!root) return null;
  if (key < root.val) root.left = deleteNode(root.left, key);
  else if (key > root.val) root.right = deleteNode(root.right, key);
  else {
    if (!root.left) return root.right;
    else if (!root.right) return root.left;
    let minNode = findMin(root.right);
    root.val = minNode.val;
    // Missing deletion of minNode here
  }
  return root;
}

function findMin(node) {
  while (node.left) node = node.left;
  return node;
}

medium

A. The function does not handle leaf nodes correctly.

B. The base case for recursion is missing.

C. The function uses incorrect comparison operators.

D. The code does not delete the in-order successor node after copying its value.

Solution

1. Step 1: Analyze deletion of node with two children

   When deleting a node with two children, we replace its value with the in-order successor's value.

2. Step 2: Check if in-order successor node is deleted

   The code copies the successor's value but does not delete the successor node itself, causing duplicate values and incorrect tree.

3. Final Answer:

   The code does not delete the in-order successor node after copying its value. -> Option D

4. Quick Check:

   Must delete successor node after value copy [OK]

Hint: Always delete successor node after copying value [OK]

Common Mistakes:

• Forgetting to delete the in-order successor node
• Missing base case (though present here)
• Incorrectly handling leaf nodes
• Using wrong comparison operators

5. You have a BST with nodes: 8, 3, 10, 1, 6, 14, 4, 7, 13.
After deleting node 3, what is the correct in-order traversal of the BST?

hard

A. [1, 4, 6, 7, 8, 10, 13, 14]

B. [1, 3, 4, 6, 7, 8, 10, 13, 14]

C. [1, 6, 7, 8, 10, 13, 14]

D. [1, 4, 7, 8, 10, 13, 14]

Solution

1. Step 1: Identify node 3 and its children

   Node 3 has two children: 1 (left) and 6 (right). 6 has children 4 and 7.

2. Step 2: Delete node 3 by replacing with in-order successor

   The in-order successor of 3 is 4 (smallest in right subtree). Replace 3 with 4, then delete original 4 node.

3. Step 3: Perform in-order traversal after deletion

   In-order traversal visits nodes in ascending order: 1, 4, 6, 7, 8, 10, 13, 14.

4. Final Answer:

   [1, 4, 6, 7, 8, 10, 13, 14] -> Option A

5. Quick Check:

   In-order after deleting 3 = [1, 4, 6, 7, 8, 10, 13, 14] [OK]

Hint: Replace deleted node with in-order successor, then traverse [OK]

Common Mistakes:

• Forgetting to delete the in-order successor node
• Using in-order predecessor instead
• Skipping nodes in traversal