BST Inorder Predecessor in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the left child of a node in a BST.

DSA Java

TreeNode leftChild = node.[1];

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

Aright

Bleft

Cparent

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' to access the left child.

Confusing 'parent' with child nodes.

The left child of a node in a BST is accessed using 'left'.

2fill in blank

medium

Complete the code to find the maximum node in the left subtree.

DSA Java

while (current.[1] != null) {
    current = current.[1];
}

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

Aright

Bleft

Cparent

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Going left instead of right to find maximum.

Stopping traversal too early.

To find the maximum node in a subtree, keep going right until null.

3fill in blank

hard

Fix the error in the code to move up to the parent node.

DSA Java

current = current.[1];

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

Aleft

Bright

Croot

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' or 'right' to move up the tree.

Using 'root' which is not a pointer in the node.

To move up in the tree, use the 'parent' pointer.

4fill in blank

hard

Fill both blanks to check if current node is a right child of its parent.

DSA Java

if (current == current.[1].[2]) {
    current = current.[1];
}

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

Aparent

Bleft

Cright

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing left and right child checks.

Not accessing the parent before checking children.

We check if current is the right child of its parent by comparing with parent's right.

5fill in blank

hard

Fill all three blanks to complete the inorder predecessor function.

DSA Java

public TreeNode inorderPredecessor(TreeNode node) {
    if (node.[1] != null) {
        TreeNode current = node.[1];
        while (current.[2] != null) {
            current = current.[2];
        }
        return current;
    }
    TreeNode current = node;
    TreeNode parent = node.[3];
    while (parent != null && current == parent.[2]) {
        current = parent;
        parent = parent.[3];
    }
    return parent;
}

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

Aleft

Bright

Cparent

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' when finding max in left subtree.

Not moving up correctly using parent pointers.

The inorder predecessor is the maximum node in left subtree or the nearest ancestor for which node is in right subtree.

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