BST Inorder Predecessor in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a pointer to the root node of a BST.

DSA C

struct Node* [1] = NULL;

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

Atemp

Bprev

Croot

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that does not represent the root node.

The root pointer is commonly named 'root' to represent the root of the BST.

2fill in blank

medium

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

DSA C

current = current->[1];

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

Anext

Bleft

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' to move left.

To move to the left child in a BST, use the 'left' pointer.

3fill in blank

hard

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

DSA C

while (temp->[1] != NULL) {
    temp = temp->right;
}

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

Aright

Bleft

Cnext

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

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

To find the rightmost node in the left subtree, move right until NULL.

4fill in blank

hard

Fill both blanks to correctly find the inorder predecessor when no left subtree exists.

DSA C

while (current->[1] != NULL && current == current->[2]->left) {
    current = current->parent;
}

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

Aparent

Bright

Cleft

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' to check child relation.

We move up using 'parent' pointer while current is the left child of its parent.

5fill in blank

hard

Fill all three blanks to complete the function that returns the inorder predecessor of a given node.

DSA C

struct Node* inorderPredecessor(struct Node* root, struct Node* current) {
    if (current->[1] != NULL) {
        struct Node* temp = current->left;
        while (temp->[2] != NULL) {
            temp = temp->right;
        }
        return temp;
    }
    struct Node* predecessor = NULL;
    while (root != NULL) {
        if (current->data > root->data) {
            predecessor = root;
            root = root->[3];
        } else {
            root = root->left;
        }
    }
    return predecessor;
}

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

Aleft

Bright

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing left and right pointers in traversal.

1) Check if left child exists (current->left). 2) Find rightmost node in left subtree (temp->right). 3) Move right in BST while searching for predecessor (root->right).

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 in BST

   Inorder traversal visits nodes in ascending order of their values.

2. Step 2: Define inorder predecessor

   The inorder predecessor of a node is the node visited just before it in inorder traversal, which means it has 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 given node node in a BST when the node has a left subtree?

easy

A. Node* pred = node->left; while (pred->right != NULL) pred = pred->right;

B. Node* pred = node->right; while (pred->left != NULL) pred = pred->left;

C. Node* pred = node->left; while (pred->left != NULL) pred = pred->left;

D. Node* pred = node->right; while (pred->right != NULL) pred = pred->right;

Solution

1. Step 1: Identify inorder predecessor in left subtree

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

2. Step 2: Trace code logic

   Starting from node->left, moving right until right child is NULL finds the rightmost node.

3. Final Answer:

   Node* pred = node->left; while (pred->right != NULL) pred = pred->right; -> Option A

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
• Moving left repeatedly instead of right
• Starting from node->right instead of node->left

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

        20
       /  \
     10    30
       \     \
       15    40

medium

A. 30

B. 15

C. 20

D. 10

Solution

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

   Node 15 has no left child, so predecessor is an 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 to root.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   Predecessor of 15 is 10 [OK]

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

Common Mistakes:

• Choosing node itself as predecessor
• Choosing larger ancestor like 20
• Confusing predecessor with successor

4. Consider this buggy C function to find the inorder predecessor of a node in a BST:

Node* inorderPredecessor(Node* root, Node* node) {
    if (node->left != NULL) {
        Node* pred = node->left;
        while (pred->left != NULL) {
            pred = pred->left;
        }
        return pred;
    }
    Node* pred = NULL;
    Node* ancestor = root;
    while (ancestor != node) {
        if (node->data > ancestor->data) {
            pred = ancestor;
            ancestor = ancestor->right;
        } else {
            ancestor = ancestor->left;
        }
    }
    return pred;
}

What is the main error in this code?

medium

A. The ancestor traversal is incorrect and may cause infinite loop

B. The function does not handle the case when node has no left subtree

C. The while loop should move right in left subtree, not left

D. The function returns NULL even if predecessor exists

Solution

1. Step 1: Analyze left subtree traversal

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

2. Step 2: Check code loop

   Code incorrectly moves left repeatedly, so it finds leftmost node instead of rightmost.

3. Final Answer:

   The while loop should move right in left subtree, not left -> Option C

4. Quick Check:

   Left subtree traversal must go right for predecessor [OK]

Hint: In left subtree, go right to find predecessor [OK]

Common Mistakes:

• Moving left instead of right in left subtree
• Ignoring ancestor traversal logic
• Assuming predecessor is always left child

5. You have a BST where each node has a pointer to its parent. Write the logic to find the inorder predecessor of a node node that has no left child. Which approach correctly finds the predecessor?

hard

A. Return the right child of the node's parent

B. Move up using parent pointers until you find a node which is right child of its parent; then return that parent

C. Move up using parent pointers until you find a node which is left child of its parent; then return that parent

D. Return the left child of the node's parent

Solution

1. Step 1: Understand no left child case

   If node has no left child, predecessor is an ancestor where node is in right subtree.

2. Step 2: Use parent pointers to find such ancestor

   Move up until current node is right child of its parent; that parent is predecessor.

3. Final Answer:

   Move up using parent pointers until you find a node which is right child of its parent; then return that parent -> Option B

4. Quick Check:

   Go up to right-child parent for predecessor [OK]

Hint: Go up until node is right child, then parent is predecessor [OK]

Common Mistakes:

• Stopping at left child parent
• Returning sibling nodes incorrectly
• Assuming predecessor is always child node