BST Inorder Predecessor in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum value node in a BST subtree.

DSA C++

Node* findMax(Node* root) {
    while (root && root->[1]) {
        root = root->right;
    }
    return root;
}

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

Aleft

Bchild

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' pointer to find maximum node.

Not checking if root is null before accessing pointers.

To find the maximum node in a BST subtree, we keep moving to the right child until there is no right child.

2fill in blank

medium

Complete the code to find the inorder predecessor when the left subtree exists.

DSA C++

Node* inorderPredecessor(Node* root) {
    if (root->left) {
        return findMax(root->[1]);
    }
    return nullptr;
}

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

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using right subtree instead of left subtree.

Returning nullptr without checking left subtree.

If the left subtree exists, the inorder predecessor is the maximum node in the left subtree.

3fill in blank

hard

Fix the error in the code to find the inorder predecessor when no left subtree exists.

DSA C++

Node* inorderPredecessor(Node* root) {
    if (!root->left) {
        Node* p = root->[1];
        while (p && root == p->left) {
            root = p;
            p = p->[1];
        }
        return p;
    }
    return findMax(root->left);
}

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

Aparent

Bchild

Cright

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using left or right pointers instead of parent to move up.

Not checking the condition in the while loop correctly.

When no left subtree exists, move up using parent pointers until you find a node which is a right child of its parent. That parent is the inorder predecessor.

4fill in blank

hard

Fill both blanks to complete the function that returns the inorder predecessor of a node in BST.

DSA C++

Node* inorderPredecessor(Node* root) {
    if (root->left) {
        return findMax(root->[1]);
    }
    Node* p = root->[2];
    while (p && root == p->left) {
        root = p;
        p = p->parent;
    }
    return p;
}

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up left and right pointers.

Using wrong pointer to move up the tree.

If left subtree exists, predecessor is max in left subtree (use left pointer). Otherwise, move up using parent pointer.

5fill in blank

hard

Fill all three blanks to complete the code that finds the inorder predecessor in BST with parent pointers.

DSA C++

Node* inorderPredecessor(Node* root) {
    if (root->[1]) {
        return findMax(root->[2]);
    }
    Node* p = root->[3];
    while (p && root == p->left) {
        root = p;
        p = p->parent;
    }
    return p;
}

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using right pointer instead of left for subtree.

Using wrong pointer to move up the tree.

The first blank checks if left subtree exists (left pointer). The second blank accesses left subtree to find max. The third blank moves up using parent pointer.

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