BST Delete Operation in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is null before deletion.

DSA Typescript

if (root === [1]) {
    return root;
}

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

Aundefined

B0

Cnull

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using undefined instead of null causes the check to fail.

Using 0 or false is incorrect because nodes are objects, not booleans or numbers.

We check if the root is null to know if the tree or subtree is empty, so we return it as is.

2fill in blank

medium

Complete the code to move left or right in the BST based on the key value.

DSA Typescript

if (key < root.val) {
    root.left = deleteNode(root.left, [1]);
} else if (key > root.val) {
    root.right = deleteNode(root.right, key);
}

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

Akey

Broot.val

Croot.left

Droot.right

Attempts:

3 left

💡 Hint

Common Mistakes

Passing root.val instead of key causes wrong recursion.

Passing root.left or root.right is incorrect because those are nodes, not keys.

We pass the key down the tree to find the node to delete.

3fill in blank

hard

Fix the error in finding the minimum value node in the right subtree.

DSA Typescript

let minNode = root.right;
while (minNode.[1] !== null) {
    minNode = minNode.left;
}

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

Aright

Bleft

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Checking minNode.right in the while condition causes infinite loop or wrong node.

Using minNode.val or minNode.parent is incorrect for traversal.

To find the minimum node in the right subtree, we keep going left until there is no left child. The loop condition checks if minNode.left is not null to continue.

4fill in blank

hard

Fill both blanks to replace the node's value with the minimum node's value and delete the minimum node.

DSA Typescript

root.val = minNode.[1];
root.right = deleteNode(root.right, [2]);

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

Aval

Bleft

CminNode.val

DminNode.left

Attempts:

3 left

💡 Hint

Common Mistakes

Using minNode.left or minNode.right instead of minNode.val causes errors.

Passing minNode.left instead of minNode.val to deleteNode causes wrong deletion.

We copy the minimum node's value to the current node, then delete the minimum node by passing its value to the delete function.

5fill in blank

hard

Fill all three blanks to handle the case when the node to delete has only one child.

DSA Typescript

if (root.left === null) {
    return root.[1];
} else if (root.right === null) {
    return root.[2];
}

// Node with two children handled later

return [3];

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

Aleft

Bright

Croot

DdeleteNode(root.left, key)

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right in returns causes wrong tree structure.

Returning deleteNode call instead of root in last blank causes infinite recursion.

If the node has no left child, we return its right child to replace it. If no right child, we return its left child. Otherwise, we return the current root after handling two children case.

Practice

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

easy

A. The entire tree is rebuilt

B. The node is replaced by its right child

C. The node is replaced by its left child

D. The node is simply removed without affecting other nodes

Solution

1. Step 1: Understand leaf node deletion

   A leaf node has no children, so deleting it means just removing it from the tree.

2. Step 2: Effect on the BST

   Since it has no children, no other nodes need to be moved or replaced.

3. Final Answer:

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

4. Quick Check:

   Leaf node deletion = simple removal [OK]

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

Common Mistakes:

• Thinking children need rearrangement
• Replacing leaf with child nodes
• Rebuilding the whole tree

2. Which of the following is the correct TypeScript function signature for deleting a node from a BST?

easy

A. function deleteNode(key: number): TreeNode

B. function deleteNode(root: number, key: TreeNode): void

C. function deleteNode(root: TreeNode | null, key: number): TreeNode | null

D. function deleteNode(root: TreeNode, key: string): TreeNode

Solution

1. Step 1: Analyze parameter types

   The root should be a TreeNode or null, and key should be a number to identify the node.

2. Step 2: Analyze return type

   The function returns the updated root node or null if tree becomes empty.

3. Final Answer:

   function deleteNode(root: TreeNode | null, key: number): TreeNode | null -> Option C

4. Quick Check:

   Correct parameter and return types = function deleteNode(root: TreeNode | null, key: number): TreeNode | null [OK]

Hint: Root can be null; key is number; returns updated root [OK]

Common Mistakes:

• Swapping parameter types
• Using wrong return type
• Using string instead of number for key

3. Given the BST below and the delete operation deleteNode(root, 10), what is the in-order traversal of the BST after deletion?

BST structure:
15
/ \
10 20
/ \ \
8 12 25

medium

A. [8, 12, 15, 20, 25]

B. [8, 10, 12, 15, 20, 25]

C. [8, 15, 20, 25]

D. [12, 15, 20, 25]

Solution

1. Step 1: Delete node 10 with two children

   Node 10 has two children (8 and 12). Replace it with its inorder successor, which is 12.

2. Step 2: Remove node 12 from its original position

   After replacement, remove node 12 from right subtree of 10.

3. Step 3: In-order traversal after deletion

   Traversal is [8, 12, 15, 20, 25].

4. Final Answer:

   [8, 12, 15, 20, 25] -> Option A

5. Quick Check:

   Inorder after deleting 10 = [8, 12, 15, 20, 25] [OK]

Hint: Replace node with inorder successor for two children [OK]

Common Mistakes:

• Not replacing node 10 correctly
• Leaving node 10 in tree
• Incorrect traversal order

4. Consider this TypeScript code snippet for deleting a node from a BST. What is the bug?

function deleteNode(root: TreeNode | null, key: number): TreeNode | null {
  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;
    if (!root.right) return root.left;
    let minNode = findMin(root.right);
    root.val = minNode.val;
    deleteNode(root.right, minNode.val);
  }
  return root;
}

medium

A. No return statement after deletion

B. Missing assignment of root.right after deleting minNode

C. Wrong comparison operators for key

D. Incorrect base case for empty root

Solution

1. Step 1: Analyze deletion of minNode

   After copying minNode's value, the code calls deleteNode on root.right but does not assign the result back to root.right.

2. Step 2: Effect of missing assignment

   Without assignment, the minNode is not removed from the right subtree, causing duplicate values.

3. Final Answer:

   Missing assignment of root.right after deleting minNode -> Option B

4. Quick Check:

   Assign updated subtree after deletion [OK]

Hint: Always assign recursive delete result back to subtree [OK]

Common Mistakes:

• Forgetting to assign after recursive delete
• Misplacing base case
• Using wrong comparison signs

5. You have a BST with nodes: 30, 20, 40, 10, 25, 35, 50. You delete node 30 (root). Which node becomes the new root after deletion using the standard BST delete operation?

hard

A. 35

B. 25

C. 40

D. 20

Solution

1. Step 1: Identify node 30's children

   Node 30 has two children: left subtree with 20 and right subtree with 40.

2. Step 2: Find inorder successor of 30

   Inorder successor is the smallest node in right subtree, which is 35.

3. Step 3: Replace 30 with 35 and delete 35 from right subtree

   35 replaces 30 as root, and 35 is removed from its original position.

4. Final Answer:

   35 -> Option A

5. Quick Check:

   Inorder successor replaces deleted node [OK]

Hint: Use inorder successor (smallest right subtree) for two-child node deletion [OK]

Common Mistakes:

• Using inorder predecessor instead
• Choosing wrong replacement node
• Not updating root after deletion