Deletion in BST in Data Structures Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the node to delete in a BST.

Data Structures Theory

if key < root.val:
    root.left = deleteNode(root.left, [1])
else:
    root.right = deleteNode(root.right, key)

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

Aroot.val

Broot.left

Ckey

Droot.right

Attempts:

3 left

💡 Hint

Common Mistakes

Using root.val instead of key in the recursive call.

Passing root.left instead of key as the argument.

To delete a node, we compare the key with the current node's value. If the key is smaller, we move to the left subtree by calling deleteNode with the key.

2fill in blank

medium

Complete the code to find the minimum value node in the right subtree during deletion.

Data Structures Theory

def minValueNode(node):
    current = node
    while current.[1] is not None:
        current = current.left
    return current

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

Aright

Bleft

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Checking current.right instead of current.left.

Returning current before the loop.

The minimum value node in a BST is found by going as far left as possible.

3fill in blank

hard

Fix the error in the code that replaces the node's value with the inorder successor's value.

Data Structures Theory

temp = minValueNode(root.[1])
root.val = temp.val
root.[2] = deleteNode(root.right, temp.val)

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using root.left instead of root.right.

Deleting from the wrong subtree.

The inorder successor is the minimum node in the right subtree, so we must look at root.right and delete from root.right.

4fill in blank

hard

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

Data Structures Theory

if root.left is None:
    return root.[1]
elif root.[2] is None:
    return root.left

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

Aright

Bleft

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Returning root.left when left is None.

Checking root.val instead of root.left or root.right.

If the left child is missing, return the right child. If the right child is missing, return the left child.

5fill in blank

hard

Fill all three blanks to complete the deletion function for a BST.

Data Structures Theory

def deleteNode(root, key):
    if root is None:
        return root
    if key < root.val:
        root.left = deleteNode(root.left, [1])
    elif key > root.val:
        root.right = deleteNode(root.right, [2])
    else:
        if root.left is None:
            return root.[3]
        elif root.right is None:
            return root.left
        temp = minValueNode(root.right)
        root.val = temp.val
        root.right = deleteNode(root.right, temp.val)
    return root

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 in recursive calls.

Returning root.left when left is None.

When searching for the node to delete, we pass the key down the tree. If the left child is missing, we return the right child.

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