BST Inorder Predecessor in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the left child of a node in a BST.

DSA C#

TreeNode leftChild = node.[1];

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

ALeft

BRight

CParent

DRoot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Right' instead of 'Left' to access the left child.

Trying to access a non-existent 'Parent' property.

The left child of a node in a BST is accessed using the 'Left' property.

2fill in blank

medium

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

DSA C#

while (current.[1] != null) {
    current = current.[1];
}

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

ARight

BLeft

CParent

DRoot

Attempts:

3 left

💡 Hint

Common Mistakes

Going left instead of right to find the maximum.

Using 'Parent' which is not always available.

The maximum node in a BST subtree is found by going as far right as possible.

3fill in blank

hard

Fix the error in the code to find the inorder predecessor when the left subtree is null.

DSA C#

while (node.[1] != null && node.[1].Value > key) {
    node = node.[1];
}

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

ALeft

BRight

CRoot

DParent

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to move left or right when left subtree is null.

Not handling the case when parent is null.

When left subtree is null, we move up to the parent to find the inorder predecessor.

4fill in blank

hard

Fill both blanks to complete the method that finds the inorder predecessor in a BST.

DSA C#

if (node.[1] != null) {
    TreeNode temp = node.[1];
    while (temp.[2] != null) {
        temp = temp.[2];
    }
    return temp;
}

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

ALeft

BRight

CParent

DRoot

Attempts:

3 left

💡 Hint

Common Mistakes

Going left twice instead of left then right.

Using parent pointer incorrectly here.

If left subtree exists, predecessor is the rightmost node in left subtree.

5fill in blank

hard

Fill all three blanks to complete the method that finds the inorder predecessor when left subtree is null.

DSA C#

TreeNode predecessor = null;
while (node != null) {
    if (key > node.Value) {
        predecessor = node;
        node = node.[1];
    } else {
        node = node.[2];
    }
}
return predecessor; // predecessor is [3]

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

ARight

BLeft

Cthe last node smaller than key

Dthe first node greater than key

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right moves.

Returning the first node greater than key instead of smaller.

When left subtree is null, move right if key is greater, else left; predecessor is last smaller node.

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.

2. Step 2: Define predecessor in inorder traversal

   The predecessor is the node visited just before the current node, so it must have the largest value smaller than the current node.

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 largest smaller value in BST [OK]

Common Mistakes:

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

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

easy

A. Node predecessor = node.Left; while (predecessor.Left != null) predecessor = predecessor.Left;

B. Node predecessor = node.Right; while (predecessor.Left != null) predecessor = predecessor.Left;

C. Node predecessor = node.Right; while (predecessor.Right != null) predecessor = predecessor.Right;

D. Node predecessor = node.Left; while (predecessor.Right != null) predecessor = predecessor.Right;

Solution

1. Step 1: Check where predecessor lies if left subtree exists

   Predecessor is the rightmost node in the left subtree.

2. Step 2: Analyze code snippets

   Node predecessor = node.Left; while (predecessor.Right != null) predecessor = predecessor.Right; starts at left child and moves right until no more right child, correctly finding predecessor.

3. Final Answer:

   Node predecessor = node.Left; while (predecessor.Right != null) predecessor = predecessor.Right; -> 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 left repeatedly instead of right in left subtree
• Starting from right child instead of left
• Ignoring the while loop to find rightmost node

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

       20
      /  \
    10    30
      \
      15

medium

A. 30

B. 10

C. 20

D. 15

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 the path.

3. Final Answer:

   10 -> Option B

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 than 15
• Choosing 15 itself
• Choosing 30 which is right subtree

4. Consider the following C# method to find the inorder predecessor of a node in a BST. What is the bug?

Node FindPredecessor(Node root, Node node) {
  if (node.Left != null) {
    Node temp = node.Left;
    while (temp.Left != null) {
      temp = temp.Left;
    }
    return temp;
  }
  Node predecessor = null;
  Node current = root;
  while (current != null) {
    if (node.Value > current.Value) {
      predecessor = current;
      current = current.Right;
    } else {
      current = current.Left;
    }
  }
  return predecessor;
}

medium

A. The while loop should check temp.Right, not temp.Left

B. The method does not handle null root

C. The method returns node itself as predecessor

D. The method incorrectly updates predecessor in the else block

Solution

1. Step 1: Analyze left subtree traversal

   Predecessor in left subtree is rightmost node, so loop should move right, not left.

2. Step 2: Identify incorrect loop condition

   Current code loops while temp.Left != null, should be temp.Right != null to find rightmost node.

3. Final Answer:

   The while loop should check temp.Right, not temp.Left -> Option A

4. Quick Check:

   Rightmost node in left subtree = predecessor [OK]

Hint: Go right in left subtree to find predecessor [OK]

Common Mistakes:

• Checking left child repeatedly instead of right
• Ignoring null root case (not critical here)
• Confusing predecessor update logic

5. You want to find the inorder predecessor of a node with value 25 in the BST below:

        40
       /  \
     20    60
    /  \  /  \
  10  30 50  70
      /  
    25   

Which node is the inorder predecessor of 25?

hard

A. 30

B. 10

C. 20

D. 25 has no predecessor

Solution

1. Step 1: Check if node 25 has left subtree

   Node 25 has no left child, so predecessor is ancestor smaller than 25.

2. Step 2: Find nearest ancestor smaller than 25

   Ancestor 20 is smaller than 25 and closest on path from root to 25.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   Predecessor of 25 is 20 [OK]

Hint: If no left child, predecessor is nearest smaller ancestor [OK]

Common Mistakes:

• Choosing 30 which is larger than 25
• Choosing 10 which is smaller but not nearest ancestor
• Thinking 25 has no predecessor