BST Delete Operation in DSA C++ - 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.

Node* deleteNode(Node* root, int key) {
    if (!root) return nullptr;
    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;
        Node* temp = findMin(root->right);
        root->val = temp->val;
        root->right = deleteNode(root->right, temp->val);
    }
    return root;
}

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls 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 minimum node in the right subtree (also a small recursion or loop).

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 height.

Input Size (n)	Approx. Operations
10	Up to 4-5 steps (tree height)
100	Up to 7-8 steps
1000	Up to 10-11 steps

Pattern observation: The steps grow slowly with input size if the tree is balanced, but can grow linearly if unbalanced.

Final Time Complexity

Time Complexity: O(h) where h is the tree height

This means the time depends on how tall the tree is, not just how many nodes it has.

Common Mistake

[X] Wrong: "Deleting a node always takes the same time regardless of tree shape."

[OK] Correct: If the tree is very unbalanced (like a linked list), deletion can take much longer because you must go through many nodes.

Interview Connect

Understanding how deletion time depends on tree height helps you explain trade-offs in tree design and shows you can analyze recursive code well.

Self-Check

"What if the BST is always perfectly balanced? How would the time complexity change then?"

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 replaced by its right child.

B. The entire tree is restructured to maintain BST properties.

C. The node is simply removed and its parent's pointer is set to null.

D. The node is replaced by its left child.

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: Update parent's pointer

   The parent's pointer to this node is set to null, effectively removing the node.

3. Final Answer:

   The node is simply removed and its parent's pointer is set to null. -> Option C

4. Quick Check:

   Leaf node deletion = remove node and nullify parent's pointer [OK]

Hint: Leaf node deletion just removes the node, no restructuring needed [OK]

Common Mistakes:

• Thinking the tree needs full restructuring
• Replacing leaf node with children when none exist
• Confusing leaf node deletion with internal node deletion

2. Which of the following is the correct syntax to delete a node with value key from a BST using a recursive function named deleteNode in C++?

easy

A. root.deleteNode(key);

B. deleteNode(root, key);

C. deleteNode(key, root);

D. root = deleteNode(root, key);

Solution

1. Step 1: Understand function signature

   The function returns the new root after deletion, so assignment is needed.

2. Step 2: Correct function call

   Calling root = deleteNode(root, key); updates the root pointer correctly.

3. Final Answer:

   root = deleteNode(root, key); -> Option D

4. Quick Check:

   Assign returned root after deletion [OK]

Hint: Always assign root = deleteNode(root, key) to update tree [OK]

Common Mistakes:

• Not assigning the returned root to update the tree
• Swapping argument order incorrectly
• Using method call syntax on root pointer

3. Consider the BST below and the code snippet that deletes the node with value 10:

    15
   /  \
  10  20
 /    / \
8    17 25

root = deleteNode(root, 10); What will be the in-order traversal of the BST after deletion?

medium

A. [8, 15, 10, 17, 20, 25]

B. [8, 15, 17, 20, 25]

C. [8, 15, 17, 20]

D. [8, 10, 15, 17, 20, 25]

Solution

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

   Node 10 has one child (8). Deleting 10 replaces it with 8.

2. Step 2: Perform in-order traversal after deletion

   In-order traversal visits nodes in ascending order: 8, 15, 17, 20, 25.

3. Final Answer:

   [8, 15, 17, 20, 25] -> Option B

4. Quick Check:

   In-order after deleting 10 = [8, 15, 17, 20, 25] [OK]

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

Common Mistakes:

• Forgetting to replace node 10 with its child 8
• Including deleted node 10 in traversal
• Misordering nodes in traversal

4. The following C++ code attempts to delete a node from a BST but has a bug. What is the bug?

Node* deleteNode(Node* root, int key) {
  if (!root) return root;
  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;
    Node* temp = root->left; // BUG HERE
    root->val = temp->val;
    root->left = deleteNode(root->left, temp->val);
  }
  return root;
}

medium

A. It uses the left child instead of the inorder successor (right subtree minimum) for replacement.

B. It does not handle the case when root is null.

C. It deletes the wrong key by comparing incorrectly.

D. It does not update the root pointer after deletion.

Solution

1. Step 1: Identify replacement node for deletion

   For nodes with two children, the inorder successor (minimum in right subtree) should replace the node.

2. Step 2: Spot incorrect replacement

   The code incorrectly uses the left child as replacement, which breaks BST properties.

3. Final Answer:

   It uses the left child instead of the inorder successor (right subtree minimum) for replacement. -> Option A

4. Quick Check:

   Use inorder successor (right subtree min) for two-child deletion [OK]

Hint: Replace with inorder successor (min right subtree), not left child [OK]

Common Mistakes:

• Using left child instead of inorder successor
• Not updating pointers after replacement
• Ignoring null root check

5. You have a BST with nodes: 30, 20, 40, 10, 25, 35, 50. You delete node 30 (root). Which node replaces the root to maintain BST properties?

hard

A. Node 35 (minimum node in right subtree)

B. Node 25 (maximum node in left subtree)

C. Node 40 (right child of root)

D. Node 20 (left child of root)

Solution

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

   Node 30 is root with two children: left subtree (20) and right subtree (40).

2. Step 2: Find inorder successor to replace root

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

3. Step 3: Replace root with inorder successor

   Node 35 replaces 30 to maintain BST properties after deletion.

4. Final Answer:

   Node 35 (minimum node in right subtree) -> Option A

5. Quick Check:

   Root replacement = inorder successor (min right subtree) [OK]

Hint: Replace deleted root with inorder successor (min right subtree) [OK]

Common Mistakes:

• Replacing with left subtree max instead of right subtree min
• Replacing with direct child without checking subtree
• Confusing inorder predecessor and successor