Deletion in BST in Data Structures Theory - Time & Space Complexity

Time Complexity: Deletion in BST

O(h)

Understanding Time Complexity

When deleting a node in a Binary Search Tree (BST), it is important to understand how the time taken grows as the tree size increases.

We want to know how the number of steps changes when the tree has more nodes.

Scenario Under Consideration

Analyze the time complexity of the following deletion process in a BST.

function deleteNode(root, key) {
  if (root == null) return null;
  if (key < root.value) {
    root.left = deleteNode(root.left, key);
  } else if (key > root.value) {
    root.right = deleteNode(root.right, key);
  } else {
    // Node found: handle 3 cases here
    // 1. No child, 2. One child, 3. Two children
  }
  return root;
}
    

This code searches for the node with the given key and deletes it by adjusting pointers depending on the node's children.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive traversal down the tree to find the node.
• How many times: At most once per level of the tree, from root to leaf.

How Execution Grows With Input

As the tree grows, the number of steps depends on the height of the tree.

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

Pattern observation: The steps grow roughly with the height of the tree, which is about log of the number of nodes if the tree is balanced.

Final Time Complexity

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

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

Common Mistake

[X] Wrong: "Deletion 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 deletion time helps you explain how tree shape affects performance, a key skill in data structure questions.

Self-Check

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

Practice

1. Which of the following is NOT a case handled during deletion in a Binary Search Tree (BST)?

easy

A. Deleting a node with two children

B. Deleting a node with one child

C. Deleting a node with three children

D. Deleting a node with no children (leaf node)

Solution

1. Step 1: Understand BST node children possibilities

   In a BST, each node can have zero, one, or two children. There is no case of three children because BST nodes have at most two children.

2. Step 2: Identify valid deletion cases

   Deletion handles nodes with zero, one, or two children. Three children is impossible, so it is not a valid case.

3. Final Answer:

   Deleting a node with three children -> Option C

4. Quick Check:

   BST nodes have max two children = Deleting node with three children [OK]

Hint: Remember BST nodes have max two children only [OK]

Common Mistakes:

• Thinking a node can have more than two children
• Confusing deletion cases with other tree types
• Ignoring the leaf node deletion case

2. Which of the following is the correct step to delete a node with two children in a BST?

easy

A. Replace the node with its parent node

B. Replace the node with its inorder predecessor (largest in left subtree)

C. Replace the node with any leaf node

D. Replace the node with its inorder successor (smallest in right subtree)

Solution

1. Step 1: Identify deletion method for two children

   When deleting a node with two children, the standard method is to replace it with its inorder successor, which is the smallest node in its right subtree.

2. Step 2: Confirm replacement correctness

   The inorder successor maintains BST properties after replacement, unlike arbitrary leaf or parent nodes.

3. Final Answer:

   Replace the node with its inorder successor (smallest in right subtree) -> Option D

4. Quick Check:

   Two children deletion uses inorder successor = Replace the node with its inorder successor (smallest in right subtree) [OK]

Hint: Use inorder successor for two-child node deletion [OK]

Common Mistakes:

• Using inorder predecessor instead of successor
• Replacing with any leaf node
• Replacing with parent node

3. Consider the BST below:

      15
     /  \
    10  20
   /    / \
  8    17 25

If we delete node 20, what will be the new right child of 15?

medium

A. 17

B. 25

C. 20

D. 15

Solution

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

   Node 20 has two children: 17 (left) and 25 (right).

2. Step 2: Find inorder successor of node 20

   The inorder successor is the smallest node in the right subtree of 20, which is 25. But since 25 has no left child, 25 is the successor.

3. Step 3: Replace node 20 with its inorder successor

   Replace 20 with 25. Then remove original 25 node (which is a leaf). The new right child of 15 becomes 25.

4. Final Answer:

   25 -> Option B

5. Quick Check:

   Inorder successor of 20 is 25 = 25 [OK]

Hint: Replace two-child node with inorder successor (smallest right) [OK]

Common Mistakes:

• Choosing 17 as successor instead of 25
• Not updating parent's child pointer
• Confusing inorder predecessor with successor

4. What is the error in the following deletion logic for a BST node with one child?

if node.left is not None:
    return node.left
else:
    return node.right

medium

A. It incorrectly returns the child without updating parent links

B. It should delete both children instead

C. It only works for leaf nodes

D. It should replace node with inorder successor

Solution

1. Step 1: Analyze deletion logic for one child

   The code returns the child node directly but does not update the parent's pointer to this child, which breaks the BST links.

2. Step 2: Identify missing link update

   Proper deletion requires updating the parent's child pointer to the returned node to maintain tree structure.

3. Final Answer:

   It incorrectly returns the child without updating parent links -> Option A

4. Quick Check:

   Missing parent link update = It incorrectly returns the child without updating parent links [OK]

Hint: Always update parent links when deleting nodes [OK]

Common Mistakes:

• Ignoring parent pointer updates
• Deleting both children unnecessarily
• Using inorder successor for one-child deletion

5. You have a BST where you want to delete the root node which has two children. The inorder successor is the immediate right child with no left child. After replacing the root with the inorder successor, what must you do next to maintain BST properties?

hard

A. Replace the inorder successor with its right child

B. Replace the inorder successor with its left child

C. Do nothing, the tree is already valid

D. Remove the inorder successor node from its original position

Solution

1. Step 1: Replace root with inorder successor

   The root is replaced by its inorder successor, which is the immediate right child with no left child.

2. Step 2: Adjust inorder successor's original position

   Since the inorder successor has no left child, replace it with its right child (which may be None) to maintain BST links.

3. Final Answer:

   Replace the inorder successor with its right child -> Option A

4. Quick Check:

   Inorder successor replaced by right child = Replace the inorder successor with its right child [OK]

Hint: Replace successor with its right child after root replacement [OK]

Common Mistakes:

• Forgetting to remove successor from original spot
• Replacing successor with left child (which doesn't exist)
• Assuming no further action needed