BST Inorder Predecessor in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the inorder predecessor value in a BST

What is the output of the code that finds the inorder predecessor of node 15 in the given BST?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def inorder_predecessor(root, key):
    predecessor = None
    current = root
    while current:
        if key <= current.val:
            current = current.left
        else:
            predecessor = current
            current = current.right
    return predecessor.val if predecessor else None

# BST construction
root = Node(20)
root.left = Node(10)
root.right = Node(30)
root.left.left = Node(5)
root.left.right = Node(15)
root.left.right.left = Node(12)
root.left.right.right = Node(17)

print(inorder_predecessor(root, 15))

A 17

B 10

C 12

D 5

💡 Hint

Remember, the inorder predecessor is the largest value smaller than the given key.

The inorder predecessor of 15 is the maximum value smaller than 15 in the BST. Node 12 is the rightmost node in the left subtree of 15, so it is the predecessor.

🧠 Conceptual

intermediate

Understanding inorder predecessor in BST

Which statement correctly describes the inorder predecessor of a node in a BST?

A It is the node with the largest value smaller than the given node's value.

B It is always the left child of the given node.

C It is the node with the smallest value larger than the given node's value.

D It is the rightmost node in the right subtree of the given node.

💡 Hint

Think about the inorder traversal order of BST nodes.

Inorder traversal visits nodes in ascending order. The predecessor is the node visited just before the given node, so it must be the largest smaller value.

🔧 Debug

advanced

Identify the error in inorder predecessor function

What error will the following code produce when trying to find the inorder predecessor of 5 in the BST?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def inorder_predecessor_buggy(root, key):
    predecessor = None
    current = root
    while current:
        if key < current.val:
            current = current.left
        elif key > current.val:
            predecessor = current
            current = current.right
        else:
            if current.left:
                temp = current.left
                while temp.right:
                    temp = temp.right
                return temp.val
            break
    return predecessor.val if predecessor else None

root = Node(20)
root.left = Node(10)
root.right = Node(30)
root.left.left = Node(5)
root.left.right = Node(15)

print(inorder_predecessor_buggy(root, 5))

A None

B 10

C 5

D AttributeError: 'NoneType' object has no attribute 'val'

💡 Hint

Check what happens when predecessor is None and you try to access its val.

If the node has no left subtree and predecessor is None, the code tries to return predecessor.val causing AttributeError.

❓ Predict Output

advanced

Output of inorder predecessor for root node

What is the output of the code when finding the inorder predecessor of the root node (20) in the BST?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def inorder_predecessor(root, key):
    predecessor = None
    current = root
    while current:
        if key <= current.val:
            current = current.left
        else:
            predecessor = current
            current = current.right
    return predecessor.val if predecessor else None

root = Node(20)
root.left = Node(10)
root.right = Node(30)
root.left.left = Node(5)
root.left.right = Node(15)

print(inorder_predecessor(root, 20))

A 15

B 10

C None

D 5

💡 Hint

The predecessor is the largest value smaller than 20.

The largest value smaller than 20 in the BST is 15, so the predecessor is 15.

🚀 Application

expert

Find inorder predecessor in BST with parent pointers

Given a BST node with a parent pointer, which approach correctly finds its inorder predecessor?

A If node has right child, predecessor is leftmost node in right subtree; else move up parents until node is left child.

B If node has left child, predecessor is rightmost node in left subtree; else move up parents until node is right child.

C Predecessor is always the left child of the node if it exists; otherwise None.

D Predecessor is the root node if node has no left child.

💡 Hint

Think about inorder traversal and how parent pointers help find predecessor.

If node has left subtree, predecessor is max in left subtree. Otherwise, move up until node is right child of parent.

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