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.

// Find inorder predecessor in BST
TreeNode InorderPredecessor(TreeNode root, TreeNode node) {
    if (node.Left != null) {
        TreeNode pred = node.Left;
        while (pred.Right != null) {
            pred = pred.Right;
        }
        return pred;
    }
    TreeNode predecessor = null;
    TreeNode current = root;
    while (current != null) {
        if (node.Val > current.Val) {
            predecessor = current;
            current = current.Right;
        } else {
            current = current.Left;
        }
    }
    return predecessor;
}
    

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

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, one path from root to a leaf or from node's left child down to rightmost leaf.

How Execution Grows With Input

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

Input Size (n)	Approx. Operations
10	Up to 4-5 steps (if tree height ~4)
100	Up to 7-8 steps (if tree height ~7)
1000	Up to 10 steps (if tree height ~10)

Pattern observation: The steps grow with the height of the tree, not directly with total nodes.

Final Time Complexity

Time Complexity: O(h)

This means the time depends on the height of the BST, which is the longest path from root to leaf.

Common Mistake

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

[OK] Correct: We only follow one path down the tree, not all nodes, so time depends on tree height, not total nodes.

Interview Connect

Understanding this helps you explain how tree shape affects search times and shows you can analyze recursive or loop-based tree operations clearly.

Self-Check

"What if the BST is balanced versus 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 largest value smaller than the given node

B. The node with the smallest value larger 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 order

   Inorder traversal visits nodes in ascending order for BST.

2. Step 2: Define predecessor in inorder traversal

   The predecessor is the node visited just before the current node, so it must have the largest value smaller than the current node.

3. Final Answer:

   The node with the largest value smaller than the given node -> Option A

4. Quick Check:

   Inorder predecessor = largest smaller value [OK]

Hint: Predecessor is largest smaller value in BST [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# code snippets correctly finds the inorder predecessor of a node with a left subtree in a BST?

easy

A. Node predecessor = node.Left; while (predecessor.Left != null) predecessor = predecessor.Left;

B. Node predecessor = node.Right; while (predecessor.Left != null) predecessor = predecessor.Left;

C. Node predecessor = node.Right; while (predecessor.Right != null) predecessor = predecessor.Right;

D. Node predecessor = node.Left; while (predecessor.Right != null) predecessor = predecessor.Right;

Solution

1. Step 1: Check where predecessor lies if left subtree exists

   Predecessor is the rightmost node in the left subtree.

2. Step 2: Analyze code snippets

   Node predecessor = node.Left; while (predecessor.Right != null) predecessor = predecessor.Right; starts at left child and moves right until no more right child, correctly finding predecessor.

3. Final Answer:

   Node predecessor = node.Left; while (predecessor.Right != null) predecessor = predecessor.Right; -> Option D

4. Quick Check:

   Left subtree's rightmost node = predecessor [OK]

Hint: Go left once, then right as far as possible [OK]

Common Mistakes:

• Going left repeatedly instead of right in left subtree
• Starting from right child instead of left
• Ignoring the while loop to find rightmost node

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

       20
      /  \
    10    30
      \
      15

medium

A. 30

B. 10

C. 20

D. 15

Solution

1. Step 1: Locate node 15 and check left subtree

   Node 15 has no left child, so predecessor is ancestor smaller than 15.

2. Step 2: Find nearest ancestor smaller than 15

   Ancestor 10 is smaller than 15 and is the closest on the path.

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Predecessor of 15 is 10 [OK]

Hint: If no left child, go up to smaller ancestor [OK]

Common Mistakes:

• Choosing 20 which is larger than 15
• Choosing 15 itself
• Choosing 30 which is right subtree

4. Consider the following C# method to find the inorder predecessor of a node in a BST. What is the bug?

Node FindPredecessor(Node root, Node node) {
  if (node.Left != null) {
    Node temp = node.Left;
    while (temp.Left != null) {
      temp = temp.Left;
    }
    return temp;
  }
  Node predecessor = null;
  Node current = root;
  while (current != null) {
    if (node.Value > current.Value) {
      predecessor = current;
      current = current.Right;
    } else {
      current = current.Left;
    }
  }
  return predecessor;
}

medium

A. The while loop should check temp.Right, not temp.Left

B. The method does not handle null root

C. The method returns node itself as predecessor

D. The method incorrectly updates predecessor in the else block

Solution

1. Step 1: Analyze left subtree traversal

   Predecessor in left subtree is rightmost node, so loop should move right, not left.

2. Step 2: Identify incorrect loop condition

   Current code loops while temp.Left != null, should be temp.Right != null to find rightmost node.

3. Final Answer:

   The while loop should check temp.Right, not temp.Left -> Option A

4. Quick Check:

   Rightmost node in left subtree = predecessor [OK]

Hint: Go right in left subtree to find predecessor [OK]

Common Mistakes:

• Checking left child repeatedly instead of right
• Ignoring null root case (not critical here)
• Confusing predecessor update logic

5. You want to find the inorder predecessor of a node with value 25 in the BST below:

        40
       /  \
     20    60
    /  \  /  \
  10  30 50  70
      /  
    25   

Which node is the inorder predecessor of 25?

hard

A. 30

B. 10

C. 20

D. 25 has no predecessor

Solution

1. Step 1: Check if node 25 has left subtree

   Node 25 has no left child, so predecessor is ancestor smaller than 25.

2. Step 2: Find nearest ancestor smaller than 25

   Ancestor 20 is smaller than 25 and closest on path from root to 25.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   Predecessor of 25 is 20 [OK]

Hint: If no left child, predecessor is nearest smaller ancestor [OK]

Common Mistakes:

• Choosing 30 which is larger than 25
• Choosing 10 which is smaller but not nearest ancestor
• Thinking 25 has no predecessor