BST Inorder Successor in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find 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'

Aparent

Bright

Cnext

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' to access the left child.

Trying to access a non-existent 'next' 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 right child of a node in a BST.

DSA C#

TreeNode rightChild = node[1];

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

Aright

Bparent

Cnext

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' to access the right child.

Trying to access a 'parent' property when looking for children.

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

3fill in blank

hard

Fix the error in the code to find the minimum node in a BST subtree.

DSA C#

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

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

Aleft

Bright

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' in the while condition.

Using 'parent' or 'next' which are not relevant here.

To find the minimum node, traverse left children until null.

4fill in blank

hard

Fill both blanks to complete the code that finds the inorder successor when the node has no right child.

DSA C#

TreeNode successor = node[1];
while (successor != null && node == successor[2]) {
    node = successor;
    successor = successor.parent;
}

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

Aparent

Bleft

Cright

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' in the comparison.

Using 'next' which is not a BST property.

We move up to the parent and check if the node is the right child to find the successor.

5fill in blank

hard

Fill all three blanks to complete the code that finds the inorder successor of a node in a BST.

DSA C#

if (node[1] != null) {
    successor = node.right;
    while (successor[2] != null) {
        successor = successor[3];
    }
} else {
    // code to find successor when no right child
}

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

Aright

Bleft

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' to check for the right child.

Using 'parent' instead of 'left' to find the minimum node.

If the node has a right child, the successor is the leftmost node in the right subtree.

Practice

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

easy

A. The node with the smallest key greater than the given node's key

B. The node with the largest key smaller than the given node's key

C. The root node of the BST

D. The left child of the given node

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits nodes in ascending order of their keys in a BST.

2. Step 2: Define inorder successor

   The inorder successor of a node is the next node visited after it in inorder traversal, which is the smallest key greater than the node's key.

3. Final Answer:

   The node with the smallest key greater than the given node's key -> Option A

4. Quick Check:

   Inorder successor = smallest greater key [OK]

Hint: Inorder successor is next bigger node in sorted order [OK]

Common Mistakes:

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

2. Which of the following C# method signatures correctly represents a function to find the inorder successor of a BST node?

easy

A. public void InorderSuccessor(TreeNode root, TreeNode p)

B. public TreeNode InorderSuccessor(TreeNode root, TreeNode p)

C. public TreeNode FindSuccessor(TreeNode p)

D. public int InorderSuccessor(int root, int p)

Solution

1. Step 1: Check parameter types for BST nodes

   The method should accept the BST root and the target node to find successor, both as TreeNode objects.

2. Step 2: Check return type

   The method returns the successor node, so return type must be TreeNode, not int or void.

3. Final Answer:

   public TreeNode InorderSuccessor(TreeNode root, TreeNode p) -> Option B

4. Quick Check:

   Correct signature uses TreeNode parameters and returns TreeNode [OK]

Hint: Successor method needs root and target node, returns TreeNode [OK]

Common Mistakes:

• Using int instead of TreeNode for parameters
• Returning void instead of TreeNode
• Missing root parameter

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

        20
       /  \
     10    30
       \     \
       15    40

TreeNode successor = InorderSuccessor(root, node15); Console.WriteLine(successor.val);

medium

A. 40

B. 30

C. 20

D. 15

Solution

1. Step 1: Locate node 15 and find its inorder successor

   Node 15 has no right child, so successor is the lowest ancestor whose left child is also ancestor of 15, which is 20.

2. Step 2: Confirm successor value

   20 is the next bigger node after 15 in inorder traversal.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   Successor of 15 is 20 [OK]

Hint: If no right child, go up to first bigger ancestor [OK]

Common Mistakes:

• Choosing right child 30 incorrectly
• Choosing node itself as successor
• Choosing 40 which is too far

4. The following code attempts to find the inorder successor of a node in a BST but returns null incorrectly. What is the bug?

public TreeNode InorderSuccessor(TreeNode root, TreeNode p) {
    if (p.right != null) {
        TreeNode curr = p.right;
        while (curr.left != null) curr = curr.left;
        return curr;
    }
    TreeNode successor = null;
    while (root != null) {
        if (p.val < root.val) {
            successor = root;
            root = root.left;
        } else {
            root = root.right;
        }
    }
    return successor;
}

medium

A. The code does not handle the case when p has no right child

B. The while loop condition should be curr.right instead of curr.left

C. The comparison p.val < root.val should be p.val <= root.val

D. The code is correct and should not return null incorrectly

Solution

1. Step 1: Analyze right subtree handling

   If p has right child, code correctly finds leftmost node in right subtree as successor.

2. Step 2: Analyze ancestor traversal

   If no right child, code moves up tree to find first ancestor greater than p, correctly updating successor.

3. Step 3: Confirm no logical errors

   Comparisons and loops are correct; code returns null only if no successor exists (p is max node).

4. Final Answer:

   The code is correct and should not return null incorrectly -> Option D

5. Quick Check:

   Code handles both cases correctly [OK]

Hint: Check both right subtree and ancestor paths for successor [OK]

Common Mistakes:

• Changing left to right in while loop incorrectly
• Using <= instead of < causing wrong ancestor selection
• Assuming code misses no-right-child case

5. You have a BST and want to find the inorder successor of the node with value 50. The BST is:

          40
         /  \
       20    60
      /  \     \
    10   30    70
                 \
                 80

If the node with value 50 does not exist in the tree, what should your inorder successor function return?

hard

A. null, because the node is not in the BST

B. The smallest node greater than 50, which is 60

C. The largest node smaller than 50, which is 40

D. The root node, which is 40

Solution

1. Step 1: Confirm node presence in BST

   Node with value 50 is not present in the BST, so no direct successor can be found.

2. Step 2: Understand successor function behavior

   Successor function requires the node to find its successor; if node not found, function should return null or indicate absence.

3. Final Answer:

   null, because the node is not in the BST -> Option A

4. Quick Check:

   Node missing means no successor found [OK]

Hint: Successor only valid if node exists in BST [OK]

Common Mistakes:

• Returning closest bigger node even if node missing
• Returning root node incorrectly
• Returning predecessor instead of null