Why BST Inorder Successor in DSA C++?

The Big Idea

Discover how to instantly find the next item in a tree without checking every node!

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 looking at every name and guessing who comes next is tiring and confusing.

The Problem

Manually searching for the next person in order means checking many names one by one, which takes a lot of time and can easily lead to mistakes, especially if the tree is big or complicated.

The Solution

The BST Inorder Successor method helps you quickly find the next person in order by using the tree's structure. It smartly moves through the tree without checking every node, saving time and avoiding errors.

Before vs After

✗ Before

Node* findSuccessor(Node* root, Node* target) {
  vector<Node*> nodes;
  inorderTraversal(root, nodes);
  for (int i = 0; i < nodes.size(); i++) {
    if (nodes[i] == target && i + 1 < nodes.size())
      return nodes[i + 1];
  }
  return nullptr;
}

✓ After

Node* findSuccessor(Node* root, Node* target) {
  if (target->right != nullptr) {
    Node* current = target->right;
    while (current->left != nullptr) {
      current = current->left;
    }
    return current;
  }
  Node* successor = nullptr;
  Node* ancestor = root;
  while (ancestor != nullptr) {
    if (target->data < ancestor->data) {
      successor = ancestor;
      ancestor = ancestor->left;
    } else if (target->data > ancestor->data) {
      ancestor = ancestor->right;
    } else {
      break;
    }
  }
  return successor;
}

What It Enables

This method lets you find the next item in order instantly, making tasks like sorting, searching, and navigating trees much faster and easier.

Real Life Example

Think of a digital phone book stored as a BST. Finding the next contact after a given name quickly helps you scroll through contacts efficiently without looking at every single entry.

Key Takeaways

Manual searching is slow and error-prone.

BST Inorder Successor uses tree structure to find the next node fast.

It improves efficiency in many tree-based tasks.

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 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 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 node with a key greater than the current node.

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++ function signatures correctly represents a function to find the inorder successor of a BST node given the root and the node? Assume Node is a struct with int key, and left/right pointers.

easy

A. Node inorderSuccessor(Node root, Node p);

B. int inorderSuccessor(Node root, int p);

C. void inorderSuccessor(Node* root, Node* p);

D. Node* inorderSuccessor(Node* root, Node* p);

Solution

1. Step 1: Analyze parameter types

   The function needs access to the BST root and the target node pointer, so pointers are required.

2. Step 2: Analyze return type

   The function should return a pointer to the successor node or nullptr if none exists.

3. Final Answer:

   Node* inorderSuccessor(Node* root, Node* p); -> Option D

4. Quick Check:

   Use pointers for tree nodes and return pointer [OK]

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

Common Mistakes:

• Passing nodes by value instead of pointer
• Returning void instead of Node*
• Using int instead of Node* for node parameters

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

        20
       /  \
     10    30
       \     \
       15    40

medium

A. 10

B. 20

C. 30

D. 40

Solution

1. Step 1: Locate node 15 and find successor

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

2. Step 2: Identify successor in BST

   20 is the parent of 15's ancestor and 15 is in left subtree of 20, so successor is 20.

3. Final Answer:

   20 -> Option B

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 40 which is too far
• Choosing 10 which is smaller

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

Node* inorderSuccessor(Node* root, Node* p) {
    if (p->right != nullptr) {
        Node* curr = p->right;
        while (curr->left != nullptr) {
            curr = curr->left;
        }
        return curr;
    }
    Node* succ = nullptr;
    while (root != nullptr) {
        if (p->key < root->key) {
            succ = root;
            root = root->left;
        } else {
            root = root->right;
        }
    }
    return succ;
}

What is the bug in this code?

medium

A. It does not check if p has a left child

B. It incorrectly updates succ when p->key == root->key

C. It does not handle the case when p is nullptr

D. It returns the wrong node when p has a right child

Solution

1. Step 1: Check null pointer dereference

   The code immediately accesses p->right and later p->key without verifying if p is nullptr, leading to a segmentation fault.

2. Step 2: Verify other logic

   Right subtree min is correctly found. Ancestor search properly sets succ only on left branches and handles equality by traversing right (null since no right child).

3. Final Answer:

   It does not handle the case when p is nullptr -> Option C

4. Quick Check:

   Missing p == nullptr check before dereference [OK]

Hint: Missing nullptr check for input node p [OK]

Common Mistakes:

• Ignoring nullptr checks for p
• Confusing left child check for successor
• Assuming bug in right subtree handling

5. You have a BST with nodes having keys: 5, 3, 7, 2, 4, 6, 8. You want to find the inorder successor of node with key 7. Which of the following is the correct approach to find it efficiently?

hard

A. If node 7 has right child, successor is leftmost node in right subtree; else, go up to first ancestor larger than 7

B. Successor is always the right child of node 7

C. Successor is the parent of node 7 regardless of children

D. Successor is the left child of node 7

Solution

1. Step 1: Check if node 7 has right child

   Node 7 has right child 8, so successor is leftmost node in right subtree, which is 8 itself.

2. Step 2: If no right child, find first ancestor larger than node

   This step is for cases where right child is absent, not applicable here.

3. Final Answer:

   If node 7 has right child, successor is leftmost node in right subtree; else, go up to first ancestor larger than 7 -> Option A

4. Quick Check:

   Right child exists, successor = leftmost right subtree [OK]

Hint: Check right subtree first; else climb ancestors [OK]

Common Mistakes:

• Assuming successor is always right child
• Ignoring ancestor climbing when no right child
• Confusing successor with parent node