Why BST Inorder Predecessor in DSA C?

The Big Idea

Discover how to instantly find the 'just before' value in a sorted tree without scanning everything!

The Scenario

Imagine you have a family tree drawn on paper, and you want to find the person who came just before someone in the family history. Doing this by scanning the whole paper every time is tiring and confusing.

The Problem

Manually searching for the previous person means checking many names one by one. This takes a lot of time and you might miss or confuse names, especially if the tree is big.

The Solution

The BST Inorder Predecessor helps you quickly find the person who comes just before another in the sorted order, without checking everyone. It uses the tree's structure to jump directly to the right place.

Before vs After

✗ Before

int findPredecessor(int arr[], int size, int value) {
    int predecessor = -1;
    for (int i = 0; i < size; i++) {
        if (arr[i] < value && arr[i] > predecessor) {
            predecessor = arr[i];
        }
    }
    return predecessor;
}

✓ After

Node* findInorderPredecessor(Node* root, int value) {
    Node* predecessor = NULL;
    Node* current = root;
    while (current != NULL) {
        if (value <= current->data) {
            current = current->left;
        } else {
            predecessor = current;
            current = current->right;
        }
    }
    return predecessor;
}

What It Enables

This concept lets you find the immediate smaller value in a sorted tree quickly and reliably, making searching and updating data efficient.

Real Life Example

In a contact list sorted by names, finding the person whose name comes just before another helps in quick navigation and suggestions.

Key Takeaways

Manual search is slow and error-prone for finding previous elements.

BST Inorder Predecessor uses tree structure for fast, direct access.

It enables efficient data retrieval in sorted order.

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