BST Delete Operation in DSA Javascript - Time & Space Complexity

Time Complexity: BST Delete Operation

O(h)

Understanding Time Complexity

We want to understand how long it takes to remove a value from a Binary Search Tree (BST).

Specifically, how the time changes as the tree grows bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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;
    if (!root.right) return root.left;
    let minLargerNode = root.right;
    while (minLargerNode.left) minLargerNode = minLargerNode.left;
    root.val = minLargerNode.val;
    root.right = deleteNode(root.right, minLargerNode.val);
  }
  return root;
}
    

This code removes a node with a given key from a BST, adjusting the tree to keep its order.

Identify Repeating Operations

• Primary operation: Recursive search down the tree to find the node to delete.
• How many times: At most once per tree level, moving left or right.
• Additional operation: Finding the smallest node in the right subtree (while loop) when deleting a node with two children.

How Execution Grows With Input

As the tree grows taller, the number of steps to find and delete a node grows roughly with the tree's height.

Input Size (n)	Approx. Operations (tree height)
10	About 3 to 4 steps
100	About 6 to 7 steps
1000	About 9 to 10 steps

Pattern observation: The steps grow slowly as the tree grows, depending on how balanced it is.

Final Time Complexity

Time Complexity: O(h)

This means the time to delete depends on the tree height, which is the number of levels from root to leaf.

Common Mistake

[X] Wrong: "Deleting a node always takes the same time no matter the tree size."

[OK] Correct: The time depends on how tall the tree is. A taller tree means more steps to find and fix nodes.

Interview Connect

Understanding how delete works in BSTs helps you show you know tree operations well, a common topic in interviews.

Self-Check

"What if the BST is always perfectly balanced? How would that affect the delete operation's time complexity?"

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