BST Inorder Predecessor in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the inorder predecessor in a Binary Search Tree (BST)?

The inorder predecessor of a node in a BST is the node that comes immediately before it in the inorder traversal. It is the largest node smaller than the given node.

beginner

How do you find the inorder predecessor if the node has a left subtree?

If the node has a left subtree, the inorder predecessor is the rightmost (maximum) node in that left subtree.

intermediate

What if the node does not have a left subtree? How to find its inorder predecessor?

If the node has no left subtree, move up the tree using parent pointers until you find a node which is the right child of its parent. That parent is the inorder predecessor.

intermediate

Why is the inorder predecessor important in BST operations?

It helps in operations like deletion of a node by replacing the node with its inorder predecessor to maintain BST properties.

beginner

In inorder traversal, what is the order of nodes visited?

Inorder traversal visits nodes in ascending order: left subtree, current node, then right subtree.

What is the inorder predecessor of a node with a left subtree in a BST?

A The leftmost node in the right subtree

B The parent node

C The root node

D The rightmost node in the left subtree

The inorder predecessor is the rightmost (maximum) node in the left subtree.

If a node has no left subtree, how do you find its inorder predecessor?

A Find the leftmost node in the right subtree

B Move up until you find a node that is the right child of its parent

C It has no inorder predecessor

D Move down to the leftmost leaf

You move up the tree until you find a node that is the right child of its parent; that parent is the inorder predecessor.

Why do we use the inorder predecessor in BST deletion?

A To find the root node

B To find the smallest node in the tree

C To replace a node with a node smaller than it to maintain BST order

D To balance the tree

Replacing a node with its inorder predecessor ensures the BST property is maintained after deletion.

In inorder traversal, which order are nodes visited?

A Left subtree, current node, right subtree

B Current node, left subtree, right subtree

C Right subtree, current node, left subtree

D Right subtree, left subtree, current node

Inorder traversal visits nodes in the order: left subtree, current node, then right subtree.

What is the inorder predecessor of the smallest node in a BST?

A No inorder predecessor (null)

B The root node

C Itself

D The rightmost node

The smallest node has no inorder predecessor because no smaller node exists.

Explain how to find the inorder predecessor of a node in a BST with and without a left subtree.

Think about the inorder traversal order and how predecessor relates to it.

Describe why the inorder predecessor is useful when deleting a node from a BST.

Consider how BST structure must remain sorted after deletion.

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