BST Inorder Predecessor in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum value node in a BST subtree.

DSA Python

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

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

Aleft

Bright

Cparent

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' will find the minimum, not maximum.

Trying to access 'parent' attribute which may not exist.

To find the maximum value node in a BST subtree, we keep moving to the right child until there is no right child.

2fill in blank

medium

Complete the code to move to the left child of a node.

DSA Python

def move_left(node):
    if node is not None:
        return node.[1]
    return None

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

Aright

Bparent

Cleft

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' will move to the wrong child.

Accessing 'value' instead of child nodes.

To move to the left child of a node, access the 'left' attribute.

3fill in blank

hard

Fix the error in the code to correctly find the inorder predecessor when the left subtree exists.

DSA Python

def inorder_predecessor(root, node):
    if node.left is not None:
        return find_max(node.[1])
    return None

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

Aright

Bparent

Cvalue

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.right will find the wrong subtree.

Using node.parent without checking if it exists.

The inorder predecessor is the maximum node in the left subtree, so we call find_max on node.left.

4fill in blank

hard

Fill both blanks to move up the tree to find the inorder predecessor when no left subtree exists.

DSA Python

def inorder_predecessor(root, node):
    if node.left is None:
        parent = node.[1]
        while parent is not None and node == parent.[2]:
            node = parent
            parent = parent.parent
        return parent
    return find_max(node.left)

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

Aparent

Bleft

Cright

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Checking node == parent.left instead of parent.right.

Not updating node and parent inside the loop.

When no left subtree exists, move up using parent pointers until the node is not the right child. So, parent is node.parent and we check node == parent.right.

5fill in blank

hard

Fill all three blanks to complete the full inorder predecessor function.

DSA Python

def inorder_predecessor(root, node):
    if node.[1] is not None:
        return find_max(node.[2])
    parent = node.[3]
    while parent is not None and node == parent.right:
        node = parent
        parent = parent.parent
    return parent

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

Aleft

Bright

Cparent

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using right instead of left for subtree check.

Not moving up correctly using parent pointers.

If node has a left child, predecessor is max in left subtree (node.left). Otherwise, move up using parent pointers.

Practice

1. What is the inorder predecessor of a node in a Binary Search Tree (BST)?

easy

A. The node with the largest value smaller than the given node

B. The node with the smallest value larger than the given node

C. The left child of the given node

D. The right child of the given node

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits nodes in ascending order for BST, so predecessor is the previous node in this order.

2. Step 2: Define inorder predecessor

   The inorder predecessor is the node with the largest value smaller than the current node's value.

3. Final Answer:

   The node with the largest value smaller than the given node -> Option A

4. Quick Check:

   Inorder predecessor = largest smaller value [OK]

Hint: Predecessor is just before current node in sorted order [OK]

Common Mistakes:

• Confusing predecessor with successor
• Thinking predecessor is always left child
• Assuming predecessor is the parent node

2. Which of the following Python code snippets correctly finds the inorder predecessor of a node with a left subtree in a BST?

easy

A. temp = node.right
while temp.right:
temp = temp.right
return temp

B. temp = node.right
while temp.left:
temp = temp.left
return temp

C. temp = node.left
while temp.left:
temp = temp.left
return temp

D. temp = node.left
while temp.right:
temp = temp.right
return temp

Solution

1. Step 1: Identify inorder predecessor logic for left subtree

   If node has left child, predecessor is the rightmost node in left subtree.

2. Step 2: Check code correctness

   Code D moves to left child, then goes right as far as possible, correctly finding predecessor.

3. Final Answer:

   Code snippet D correctly finds the inorder predecessor -> Option D

4. Quick Check:

   Left subtree's rightmost node = predecessor [OK]

Hint: Go left once, then right as far as possible [OK]

Common Mistakes:

• Going right subtree instead of left
• Going left repeatedly instead of right in left subtree
• Returning left child directly without traversal

3. Given the BST below and the node with value 15, what is the inorder predecessor?

       20
      /  \
    10    30
      \     \
      15    40

medium

A. 20

B. 15

C. 10

D. None

Solution

1. Step 1: Locate node 15 and check left subtree

   Node 15 has no left child, so predecessor is ancestor smaller than 15.

2. Step 2: Find nearest ancestor smaller than 15

   Ancestor 10 is smaller than 15 and is the closest on path to root.

3. Final Answer:

   10 -> Option C

4. Quick Check:

   Predecessor of 15 is 10 [OK]

Hint: If no left child, go up to smaller ancestor [OK]

Common Mistakes:

• Choosing 20 which is larger
• Choosing node itself as predecessor
• Assuming predecessor is None without checking ancestors

4. The following code attempts to find the inorder predecessor of a node in a BST but has a bug. What is the bug?

def inorder_predecessor(root, node):
    if node.left:
        temp = node.left
        while temp.left:
            temp = temp.left
        return temp
    else:
        predecessor = None
        while root:
            if node.val > root.val:
                predecessor = root
                root = root.right
            else:
                root = root.left
        return predecessor

medium

A. The else block incorrectly updates predecessor

B. The while loop should check temp.right, not temp.left

C. The function does not handle empty tree

D. The function should return None if node has no left child

Solution

1. Step 1: Analyze left subtree traversal

   In left subtree, predecessor is rightmost node, so loop should go right, not left.

2. Step 2: Identify correct traversal direction

   Changing while temp.left to while temp.right fixes the bug.

3. Final Answer:

   The while loop should check temp.right, not temp.left -> Option B

4. Quick Check:

   Predecessor in left subtree = rightmost node [OK]

Hint: Go right in left subtree, not left [OK]

Common Mistakes:

• Looping left instead of right in left subtree
• Ignoring else block logic
• Not returning predecessor correctly

5. You want to find the inorder predecessor of a node in a BST that does NOT have a left child. Which approach correctly finds the predecessor?

hard

A. Traverse from root, track the last node smaller than current node

B. Return the rightmost node in the left subtree

C. Return the parent node if it is smaller than the current node

D. Return the left child of the current node

Solution

1. Step 1: Understand no left child case

   If node has no left child, predecessor is an ancestor smaller than node.

2. Step 2: Use root traversal to track predecessor

   Traverse from root, update predecessor when node value is greater than root's value.

3. Step 3: Return last tracked smaller ancestor

   This gives the inorder predecessor correctly.

4. Final Answer:

   Traverse from root, track the last node smaller than current node -> Option A

5. Quick Check:

   No left child -> track smaller ancestor [OK]

Hint: Track last smaller ancestor from root [OK]

Common Mistakes:

• Returning parent without checking value
• Returning left child when none exists
• Trying to find predecessor in left subtree when none exists