Why BST Inorder Successor in DSA C?

The Big Idea

Discover how to jump to the next item in a sorted tree without searching everything!

The Scenario

Imagine you have a family tree drawn on paper, and you want to find the next person in order after a given family member. Doing this by hand means scanning the entire tree every time, which is slow and confusing.

The Problem

Manually searching for the next person in order means checking many family members one by one. This is slow and easy to make mistakes, especially if the tree is big or unbalanced.

The Solution

The BST Inorder Successor method helps you quickly find the next person in order by using the tree's structure. It avoids scanning the whole tree and finds the answer by following clear steps.

Before vs After

✗ Before

int findSuccessor(Node* root, int key) {
    int successor = -1;
    inorderTraversal(root, &successor, key);
    return successor;
}

// This scans the whole tree every time.

✓ After

Node* findInorderSuccessor(Node* root, Node* target) {
    if (target->right != NULL) {
        return findMin(target->right);
    }
    Node* successor = NULL;
    Node* current = root;
    while (current != NULL) {
        if (target->data < current->data) {
            successor = current;
            current = current->left;
        } else {
            current = current->right;
        }
    }
    return successor;
}

What It Enables

This concept enables fast and reliable finding of the next item in sorted order without checking every node.

Real Life Example

In a contact list sorted by names, finding the next contact after a given name quickly helps you jump through the list without scrolling all contacts.

Key Takeaways

Manual search is slow and error-prone for finding next in order.

BST Inorder Successor uses tree structure for efficient searching.

It helps quickly find the next bigger value in a sorted tree.

Practice

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

easy

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

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

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 keys.

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 node greater than it.

3. Final Answer:

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

4. Quick Check:

   Inorder successor = next bigger node [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 left child

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

easy

A. struct Node* inorderSuccessor(struct Node* root, struct Node* n);

B. int inorderSuccessor(int root, int n);

C. void inorderSuccessor(struct Node* root);

D. struct Node inorderSuccessor(struct Node n);

Solution

1. Step 1: Analyze function parameters

   The function needs the BST root and the node to find successor for, both as pointers.

2. Step 2: Analyze return type

   It should return a pointer to the successor node, so return type is struct Node*.

3. Final Answer:

   struct Node* inorderSuccessor(struct Node* root, struct Node* n); -> Option A

4. Quick Check:

   Correct signature uses pointers for nodes [OK]

Hint: Use pointers for nodes and return pointer to successor [OK]

Common Mistakes:

• Using int instead of struct Node*
• Missing node parameter
• Returning struct Node by value

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

        20
       /  \
     10    30
       \     \
       15    40

medium

A. 10

B. 30

C. 20

D. 40

Solution

1. Step 1: Locate node 15 and find its right subtree

   Node 15 has no right child, so successor is an ancestor.

2. Step 2: Move up to find the lowest ancestor greater than 15

   Parent 10 is less than 15, move up to 20 which is greater than 15.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   Next bigger after 15 is 20 [OK]

Hint: If no right child, successor is nearest bigger ancestor [OK]

Common Mistakes:

• Choosing right child 30 incorrectly
• Choosing smaller ancestor 10
• Confusing successor with predecessor

4. Consider this buggy C code snippet to find the inorder successor of a node in BST:

struct Node* inorderSuccessor(struct Node* root, struct Node* n) {
    if (n->right != NULL) {
        struct Node* temp = n->right;
        while (temp->left != NULL) {
            temp = temp->left;
        }
        return temp;
    }
    struct Node* succ = NULL;
    while (root != NULL) {
        if (n->data < root->data) {
            succ = root;
            root = root->left;
        } else {
            root = root->right;
        }
    }
    return succ;
}

What is the main logical error in this code?

medium

A. It incorrectly updates successor when moving right

B. It returns the wrong node when node has right child

C. It uses a while loop instead of recursion

D. It does not handle the case when node has no right child

Solution

1. Step 1: Check handling of right child

   Code correctly finds leftmost node in right subtree if right child exists.

2. Step 2: Check updating successor when moving right

   In the else branch (n->data >= root->data), it incorrectly sets succ = root before moving right. Successor should NOT be updated when going right, as n is in (or at) the right subtree.

3. Step 3: Identify error

   The main logical error is incorrectly updating the successor when traversing right.

4. Final Answer:

   It incorrectly updates successor when moving right -> Option A

5. Quick Check:

   Successor updates only on left moves [OK]

Hint: Successor updates only when moving left in BST [OK]

Common Mistakes:

• Thinking recursion is required
• Updating successor when moving right
• Ignoring right subtree case

5. You have a BST and want to find the inorder successor of the node with value 25. The BST is:

         20
        /  \
      10    30
     /  \  /  \
    5  15 25  40

If the node with value 25 has no right child, what is the inorder successor and why?

hard

A. 20, because it is the root

B. 40, because it is the right child of 30

C. No successor, because 25 has no right child

D. 30, because it is the lowest ancestor greater than 25

Solution

1. Step 1: Check if node 25 has right child

   Node 25 has no right child, so successor is an ancestor.

2. Step 2: Find lowest ancestor greater than 25

   Ancestor 30 is greater than 25 and is the lowest such ancestor.

3. Final Answer:

   30, because it is the lowest ancestor greater than 25 -> Option D

4. Quick Check:

   Successor = nearest bigger ancestor [OK]

Hint: If no right child, successor is nearest bigger ancestor [OK]

Common Mistakes:

• Choosing right child 40 incorrectly
• Saying no successor exists
• Choosing root 20 incorrectly