BST Inorder Predecessor in DSA C++ - Time & Space Complexity

Time Complexity: BST Inorder Predecessor

O(h)

Understanding Time Complexity

We want to understand how long it takes to find the inorder predecessor in a Binary Search Tree (BST).

This helps us know how the search time grows as the tree gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the inorder predecessor of a given node in a BST by checking left subtree or ancestors.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing down the tree nodes using while loops.
• How many times: At most, it moves down the height of the tree twice: once to left subtree's rightmost node or once up the ancestors.

How Execution Grows With Input

As the tree grows taller, the number of steps to find the predecessor grows roughly with the tree height.

Input Size (n)	Approx. Operations
10	About 3 to 4 steps (tree height ~3)
100	About 6 to 7 steps (tree height ~7)
1000	About 9 to 10 steps (tree height ~10)

Pattern observation: The steps grow slowly, roughly proportional to the tree height, not the total nodes.

Final Time Complexity

Time Complexity: O(h)

This means the time to find the inorder predecessor grows with the height of the BST, not the total number of nodes.

Common Mistake

[X] Wrong: "Finding the inorder predecessor always takes O(n) time because we might check all nodes."

[OK] Correct: We only follow a path down the tree, never visiting all nodes. So the time depends on tree height, not total nodes.

Interview Connect

Understanding this helps you explain how tree structure affects search times, a key skill in many coding problems.

Self-Check

"What if the BST is balanced vs completely skewed? How would the time complexity change?"

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