BST Inorder Predecessor in DSA Java - 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).

Specifically, how the time grows as the tree gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Find inorder predecessor in BST
Node inorderPredecessor(Node root, Node target) {
  if (target.left != null) {
    Node curr = target.left;
    while (curr.right != null) {
      curr = curr.right;
    }
    return curr;
  }
  Node predecessor = null;
  Node curr = root;
  while (curr != null) {
    if (target.data > curr.data) {
      predecessor = curr;
      curr = curr.right;
    } else {
      curr = curr.left;
    }
  }
  return predecessor;
}
    

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

Identify Repeating Operations

Look for loops or repeated steps in the code.

• Primary operation: Traversing down the tree nodes using while loops.
• How many times: At most, the number of nodes along a path from root to leaf or target's left subtree height.

How Execution Grows With Input

As the BST grows, the longest path from root to leaf increases.

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

Pattern observation: The steps grow roughly with the height of the tree, not 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 time proportional to the total number of nodes (n)."

[OK] Correct: The search only follows a path down the tree, not all nodes, so it depends on the tree height, not total nodes.

Interview Connect

Understanding this helps you explain how tree shape affects search times, a key skill for working with BSTs and related data structures.

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, so predecessor is the previous node in this order.

2. Step 2: Define inorder predecessor

   The inorder predecessor is the node with the largest value smaller than the current node's value.

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 just before current node 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 Java method signatures correctly represents a function to find the inorder predecessor in a BST node?

easy

A. public Node findInorderSuccessor(Node root, int key)

B. public int findInorderPredecessor(int[] arr, int key)

C. public Node findInorderPredecessor(Node root, int key)

D. public void findInorderPredecessor(Node root)

Solution

1. Step 1: Identify correct parameters for BST predecessor method

   We need the BST root and the key node value to find predecessor, so method should accept Node root and int key.

2. Step 2: Check method return type and name

   Return type should be Node (the predecessor node), and method name should indicate predecessor, so 'findInorderPredecessor(Node root, int key)' fits.

3. Final Answer:

   public Node findInorderPredecessor(Node root, int key) -> Option C

4. Quick Check:

   Correct signature includes root, key, returns Node [OK]

Hint: Method needs root and key, returns Node predecessor [OK]

Common Mistakes:

• Using array instead of BST Node
• Confusing predecessor with successor method name
• Wrong return type like void or int

3. Given the BST below and the code snippet, what will be the output?

    20
   /  \
 10   30
   
Node predecessor = findInorderPredecessor(root, 20);
System.out.println(predecessor != null ? predecessor.data : "null");

Assuming findInorderPredecessor is correctly implemented.

medium

A. null

B. 10

C. 30

D. 20

Solution

1. Step 1: Identify inorder predecessor of node 20 in BST

   Inorder predecessor is the largest node smaller than 20. Left subtree of 20 has node 10, which is smaller and largest in left subtree.

2. Step 2: Confirm output of print statement

   Predecessor node data is 10, so printed output is "10".

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Largest smaller node than 20 is 10 [OK]

Hint: Predecessor is max in left subtree if left exists [OK]

Common Mistakes:

• Printing null when left subtree exists
• Confusing successor with predecessor
• Printing the node itself instead of predecessor

4. Consider this buggy Java method to find the inorder predecessor in a BST:

public Node findInorderPredecessor(Node root, int key) {
    Node predecessor = null;
    while (root != null) {
        if (key < root.data) {
            root = root.left;
        } else {
            predecessor = root;
            root = root.right;
        }
    }
    return predecessor;
}

What is the main issue with this code?

medium

A. It does not handle the case when the node has a left subtree

B. It incorrectly updates predecessor when key equals root.data

C. It causes infinite loop for some inputs

D. It returns the successor instead of predecessor

Solution

1. Step 1: Analyze traversal logic for predecessor

   The code moves left if key < root.data, else moves right and updates predecessor. But it never checks left subtree of the node itself.

2. Step 2: Identify missing left subtree handling

   If the node has a left subtree, the predecessor is the maximum node in that left subtree, which this code does not check.

3. Final Answer:

   It does not handle the case when the node has a left subtree -> Option A

4. Quick Check:

   Missing left subtree max check causes wrong predecessor [OK]

Hint: Check left subtree max for predecessor if it exists [OK]

Common Mistakes:

• Ignoring left subtree max node
• Confusing predecessor with successor logic
• Assuming predecessor is always ancestor

5. You have a BST and want to find the inorder predecessor of a node with value 15. The node has no left child. Which approach correctly finds the predecessor?

hard

A. Find the maximum node in the left subtree of 15

B. Return the right child of node 15

C. Return null because no left child exists

D. Traverse from root, keep track of last node with value less than 15

Solution

1. Step 1: Understand predecessor when no left child

   If node has no left child, predecessor is one of its ancestors with value less than node's value.

2. Step 2: Use traversal from root to track predecessor

   Traverse from root to node 15, keep updating predecessor to last node with value less than 15.

3. Final Answer:

   Traverse from root, keep track of last node with value less than 15 -> Option D

4. Quick Check:

   No left child means ancestor with smaller value is predecessor [OK]

Hint: Track last smaller ancestor when no left child [OK]

Common Mistakes:

• Returning null immediately if no left child
• Trying to find max in non-existent left subtree
• Returning right child mistakenly