BST Inorder Predecessor in DSA Java - Execution Trace

Concept Flow - BST Inorder Predecessor

Start at root node

↓

Search for target node

↓

Found target node?

No→Return null (not found)

Yes↓

Check if left child exists

↓

Go to left subtree

↓

Find rightmost node in left subtree

↓

This node is inorder predecessor

↓

Return predecessor value

Start from root, find the target node. If it has a left child, go left then rightmost to find predecessor. Otherwise, predecessor is ancestor tracked during search.

Execution Sample

DSA Java

class Node {
  int val;
  Node left, right;
}

Node inorderPredecessor(Node root, int key) {
  // find predecessor
}

Find the inorder predecessor of a given key in a BST.

Execution Table

Step	Operation	Current Node	Pointer Changes	Visual State
1	Start at root	20	root -> 20	20 / \ 10 30
2	Compare key=10 with current=20	20	current -> 20	20 / \ 10 30
3	Go left (key < current)	10	current -> 10	20 / \ 10 30
4	Found target node 10	10	target -> 10	20 / \ 10 30
5	Check left child of 10	10	left child -> 5	20 / \ 10 30 / 5
6	Go to left child 5	5	current -> 5	20 / \ 10 30 / 5
7	Find rightmost in left subtree	5	right child -> null	20 / \ 10 30 / 5
8	Rightmost node in left subtree is 5	5	predecessor -> 5	20 / \ 10 30 / 5
9	Return predecessor value 5	5	return 5	20 / \ 10 30 / 5
10	End	5	-	Predecessor found: 5

💡 Predecessor found as rightmost node in left subtree of target node 10.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 5	After Step 7	Final
current	20	20	10	10	5	5
target	null	null	null	10	10	10
predecessor	null	null	null	null	5	5

Key Moments - 3 Insights

Why do we look for the rightmost node in the left subtree?

Because the inorder predecessor is the largest node smaller than the target, which is the rightmost node in its left subtree (see steps 6-8 in execution_table).

What if the target node has no left child?

Then the predecessor is one of the ancestors tracked during search, but this example focuses on the left subtree case (not shown here).

Why do we stop at the rightmost node in the left subtree?

Because moving right as far as possible in left subtree finds the maximum smaller value, which is the predecessor (step 7 in execution_table).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'current' at Step 3?

A 10

B 20

C 5

D 30

💡 Hint

Check the 'current' column at Step 3 in execution_table.

At which step do we identify the inorder predecessor node?

A Step 5

B Step 7

C Step 8

D Step 9

💡 Hint

Look for when 'predecessor' pointer is assigned in execution_table.

If the target node had no left child, how would the execution_table change?

A We would skip steps 5 to 8 and find predecessor among ancestors.

B We would still go to left child and rightmost node.

C We would return null immediately.

D We would search the right subtree instead.

💡 Hint

Refer to key_moments about no left child case.

Concept Snapshot

BST Inorder Predecessor:
- Find target node in BST.
- If target has left child, predecessor is rightmost node in left subtree.
- Otherwise, predecessor is nearest ancestor smaller than target.
- Used in deletion and traversal algorithms.
- Time: O(h), h = tree height.

Full Transcript

To find the inorder predecessor in a BST, start at the root and search for the target node. Once found, check if it has a left child. If yes, go to the left child and then keep moving right until no more right child exists. This rightmost node in the left subtree is the inorder predecessor. If no left child exists, the predecessor is one of the ancestors smaller than the target, tracked during the search. This process takes time proportional to the tree height.

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