BST Inorder Predecessor in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Inorder Predecessor Function in BST

What is the output of the following C++ code that finds the inorder predecessor of a given node in a BST?

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
};

Node* inorderPredecessor(Node* root, Node* p) {
    Node* predecessor = nullptr;
    while (root) {
        if (p->data <= root->data) {
            root = root->left;
        } else {
            predecessor = root;
            root = root->right;
        }
    }
    return predecessor;
}

int main() {
    Node n1{20, nullptr, nullptr};
    Node n2{10, nullptr, nullptr};
    Node n3{30, nullptr, nullptr};
    Node n4{5, nullptr, nullptr};
    Node n5{15, nullptr, nullptr};

    n1.left = &n2; n1.right = &n3;
    n2.left = &n4; n2.right = &n5;

    Node* pred = inorderPredecessor(&n1, &n5);
    if (pred) {
        std::cout << pred->data << std::endl;
    } else {
        std::cout << "No predecessor" << std::endl;
    }
    return 0;
}

A 20

B 15

C 10

D No predecessor

💡 Hint

Think about the largest value smaller than the node 15 in the BST.

The inorder predecessor of node 15 is the largest node smaller than 15. In this BST, 10 is the predecessor because it is less than 15 and the closest smaller value.

🧠 Conceptual

intermediate

Understanding Inorder Predecessor in BST

In a Binary Search Tree (BST), which of the following best describes the inorder predecessor of a node?

A The node with the largest value smaller than the given node's value

B The node with the smallest value larger than the given node's value

C The left child of the given node

D The right child of the given node

💡 Hint

Inorder traversal visits nodes in ascending order.

Inorder predecessor is the node that comes immediately before the given node in inorder traversal, which means it has the largest value smaller than the given node.

🔧 Debug

advanced

Identify the Error in Inorder Predecessor Code

What error does the following C++ code produce when trying to find the inorder predecessor of a node in a BST?

DSA C++

Node* inorderPredecessor(Node* root, Node* p) {
    if (!root) return nullptr;
    if (p->left) {
        Node* temp = p->left;
        while (temp->right) temp = temp->right;
        return temp;
    }
    Node* predecessor = nullptr;
    while (root) {
        if (p->data < root->data) {
            root = root->left;
        } else if (p->data > root->data) {
            predecessor = root;
            root = root->right;
        }
    }
    return predecessor;
}

// Called with p = nullptr

A No error, returns nullptr

B Compilation error due to missing return statement

C Infinite loop in while loop

D Segmentation fault due to dereferencing nullptr

💡 Hint

Check what happens if p is nullptr.

If p is nullptr, accessing p->left causes a segmentation fault because it dereferences a null pointer.

❓ Predict Output

advanced

Output of Inorder Predecessor for Root Node

What is the output of the following code when finding the inorder predecessor of the root node in a BST?

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
};

Node* inorderPredecessor(Node* root, Node* p) {
    if (p->left) {
        Node* temp = p->left;
        while (temp->right) temp = temp->right;
        return temp;
    }
    Node* predecessor = nullptr;
    while (root) {
        if (p->data <= root->data) {
            root = root->left;
        } else {
            predecessor = root;
            root = root->right;
        }
    }
    return predecessor;
}

int main() {
    Node n1{20, nullptr, nullptr};
    Node n2{10, nullptr, nullptr};
    Node n3{30, nullptr, nullptr};
    n1.left = &n2; n1.right = &n3;

    Node* pred = inorderPredecessor(&n1, &n1);
    if (pred) {
        std::cout << pred->data << std::endl;
    } else {
        std::cout << "No predecessor" << std::endl;
    }
    return 0;
}

A 30

B 10

C No predecessor

D 20

💡 Hint

Check if root has a left subtree and find its rightmost node.

The root node 20 has a left child 10 with no right child, so the inorder predecessor is 10.

🚀 Application

expert

Number of Nodes with Inorder Predecessor in BST

Given a BST with nodes 5, 10, 15, 20, 25 inserted in that order, how many nodes have a valid inorder predecessor?

A 4

B 5

C 3

D 2

💡 Hint

The smallest node has no predecessor.

Inorder predecessor exists for all nodes except the smallest one. Here, 5 is smallest, so 4 nodes have predecessors.

Practice

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

easy

A. The node with the smallest value larger than the given node

B. The node with the largest value smaller 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 B

4. Quick Check:

   Inorder predecessor = largest smaller value [OK]

Hint: Predecessor is just before in sorted order [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++ function signatures correctly represents a function to find the inorder predecessor of a BST node given its pointer Node* root and Node* target?

easy

A. Node* findInorderPredecessor(Node* root, Node* target);

B. void findInorderPredecessor(Node* root, int target);

C. int findInorderPredecessor(Node* root, Node* target);

D. Node findInorderPredecessor(Node root, Node target);

Solution

1. Step 1: Analyze parameter types

   The function needs the root pointer and the target node pointer to find the predecessor.

2. Step 2: Analyze return type

   The predecessor is a node pointer, so return type should be Node*.

3. Final Answer:

   Node* findInorderPredecessor(Node* root, Node* target); -> Option A

4. Quick Check:

   Correct parameters and return type [OK]

Hint: Return pointer to predecessor node [OK]

Common Mistakes:

• Using int return type instead of Node*
• Passing target as int instead of Node*
• Passing nodes by value instead of pointer

3. Consider the following BST and code snippet to find the inorder predecessor of node with value 15:

struct Node {
  int val;
  Node* left;
  Node* right;
};

Node* findMax(Node* root) {
  while (root->right) root = root->right;
  return root;
}

Node* inorderPredecessor(Node* root, Node* target) {
  if (target->left) return findMax(target->left);
  Node* predecessor = nullptr;
  while (root) {
    if (target->val > root->val) {
      predecessor = root;
      root = root->right;
    } else if (target->val < root->val) {
      root = root->left;
    } else break;
  }
  return predecessor;
}

// BST structure:
//        20
//       /  \
//     10    30
//       \     \
//       15    40

Node* root = ...; // root of above BST
Node* target = root->left->right; // node with value 15
Node* pred = inorderPredecessor(root, target);
std::cout << pred->val << std::endl;

What is the output?

medium

A. 10

B. 15

C. 20

D. 30

Solution

1. Step 1: Check if target has left subtree

   Node 15 has no left child, so we skip findMax on left subtree.

2. Step 2: Traverse from root to find predecessor

   Start at root (20). Since 15 < 20, go left to 10. Since 15 > 10, set predecessor = 10 and go right to 15. Found target, stop.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Largest smaller value than 15 is 10 [OK]

Hint: If no left child, track last smaller ancestor [OK]

Common Mistakes:

• Returning target node itself
• Returning root node 20 incorrectly
• Confusing predecessor with successor

4. The following code attempts to find the inorder predecessor of a node in a BST but has a bug. Identify the bug.

Node* inorderPredecessor(Node* root, Node* target) {
  if (target->left) {
    Node* temp = target->left;
    while (temp->left) temp = temp->left;
    return temp;
  }
  Node* predecessor = nullptr;
  while (root) {
    if (target->val > root->val) {
      predecessor = root;
      root = root->right;
    } else if (target->val < root->val) {
      root = root->left;
    } else break;
  }
  return predecessor;
}

medium

A. The predecessor variable is not initialized

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

C. The function returns nullptr always

D. The loop to find max in left subtree incorrectly goes left instead of right

Solution

1. Step 1: Check left subtree max search

   To find inorder predecessor in left subtree, we must find the maximum node, which is the rightmost node, but code goes left repeatedly.

2. Step 2: Confirm other parts are correct

   The rest of the code correctly handles cases without left subtree and initializes predecessor.

3. Final Answer:

   The loop to find max in left subtree incorrectly goes left instead of right -> Option D

4. Quick Check:

   Max in left subtree requires right traversal [OK]

Hint: Max in left subtree means go right, not left [OK]

Common Mistakes:

• Going left instead of right to find max
• Not checking if target->left is null
• Confusing predecessor with successor

5. Given a BST with nodes inserted in this order: 25, 15, 50, 10, 22, 35, 70, 4, 12, 18, 24, 31, 44, 66, 90. Write a function to find the inorder predecessor of a node with value 31. What will be the predecessor's value? Hint: Use the BST property and consider both cases when the node has a left subtree or not.

hard

A. 25

B. 35

C. 24

D. 22

Solution

1. Step 1: Locate node 31

   Node 31 is the left child of 35 and has no left subtree.

2. Step 2: Traverse from root to find predecessor

   Start at root 25 (31 > 25, predecessor=25, go right to 50). 31 < 50, go left to 35. 31 < 35, go left to 31. Found target, predecessor=25.

3. Step 3: Verify with inorder traversal

   Inorder traversal sequence around 31 is ... 24, 25, 31, 35 ... The predecessor of 31 is the largest smaller value, which is 24 (left subtree of 25).

4. Final Answer:

   24 -> Option C

5. Quick Check:

   Inorder predecessor of 31 is 24 [OK]

Hint: If no left child, predecessor is last smaller ancestor or max in left subtree [OK]

Common Mistakes:

• Choosing 25 instead of 24
• Choosing successor 35 instead of predecessor
• Ignoring left subtree check