BST Inorder Successor 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'

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using right child instead of left

Using parent instead of child

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

2fill in blank

medium

Complete the code to find the minimum node in a BST subtree.

DSA Java

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

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

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Going right instead of left

Checking parent instead of child

To find the minimum node, keep going left until no left child exists.

3fill in blank

hard

Fix the error in the code to find the inorder successor when the node has no right child.

DSA Java

TreeNode successor = null;
TreeNode ancestor = root;
while (ancestor != node) {
    if (node.val < ancestor.val) {
        successor = ancestor;
        ancestor = ancestor.[1];
    } else {
        ancestor = ancestor.right;
    }
}
return successor;

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

Achild

Bparent

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Going right instead of left

Using parent pointer incorrectly

When moving down the tree, if node's value is less than ancestor's, go left to find successor.

4fill in blank

hard

Fill both blanks to complete the method that finds the inorder successor in a BST.

DSA Java

if (node.[1] != null) {
    return findMin(node.[2]);
}

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

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using left child instead of right

Using parent instead of child

If the node has a right child, the successor is the minimum node in the right subtree.

5fill in blank

hard

Fill all three blanks to complete the findMin method and the inorderSuccessor method logic.

DSA Java

private TreeNode findMin(TreeNode node) {
    while (node.[1] != null) {
        node = node.[2];
    }
    return node;
}

public TreeNode inorderSuccessor(TreeNode root, TreeNode node) {
    if (node.[3] != null) {
        return findMin(node.right);
    }
    // Additional logic omitted for brevity
    return null;
}

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Going right in findMin

Checking left child in successor method

The findMin method goes left until no left child exists. The inorderSuccessor checks if the node has a right child.

Practice

1. What is the inorder successor of a node in a Binary Search Tree (BST)?

easy

A. The node with the smallest key greater than the given node's key

B. The node with the largest key smaller than the given node's key

C. The root node of the BST

D. The left child of the given node

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits nodes in ascending order of their keys in a BST.

2. Step 2: Define inorder successor

   The inorder successor of a node is the next node visited after it in inorder traversal, which is the smallest node greater than it.

3. Final Answer:

   The node with the smallest key greater than the given node's key -> Option A

4. Quick Check:

   Inorder successor = smallest greater key [OK]

Hint: Inorder successor is next bigger node in sorted order [OK]

Common Mistakes:

• Confusing successor with predecessor
• Thinking successor is always right child
• Assuming successor is parent node

2. Which of the following Java method signatures correctly represents a function to find the inorder successor in a BST node class?

easy

A. public void inorderSuccessor(Node p)

B. public Node inorderSuccessor(Node root, Node p)

C. public int inorderSuccessor(Node root)

D. public Node inorderSuccessor(int key)

Solution

1. Step 1: Identify required inputs and outputs

   To find inorder successor, we need the BST root and the target node; output is the successor node.

2. Step 2: Match method signature

   public Node inorderSuccessor(Node root, Node p) takes root and node, returns Node, matching requirements.

3. Final Answer:

   public Node inorderSuccessor(Node root, Node p) -> Option B

4. Quick Check:

   Input root and node, output node [OK]

Hint: Method needs root and target node, returns successor node [OK]

Common Mistakes:

• Missing root parameter
• Returning int instead of Node
• Using void return type

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

       20
      /  \
    10    30
      \
      15

medium

A. 20

B. 15

C. 30

D. 10

Solution

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

   Node 15 has no right child, so successor is ancestor with greater value.

2. Step 2: Find lowest ancestor greater than 15

   Ancestor 20 is the next bigger node after 15.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Next bigger after 15 is 20 [OK]

Hint: If no right child, successor is nearest bigger ancestor [OK]

Common Mistakes:

• Choosing 30 which is too big
• Choosing 10 which is smaller
• Choosing the node itself

4. Consider this buggy Java code snippet to find the inorder successor of a node in BST:

public Node inorderSuccessor(Node root, Node p) {
    if (p.right != null) {
        Node curr = p.right;
        while (curr.left != null) {
            curr = curr.left;
        }
        return curr;
    }
    Node succ = null;
    while (root != null) {
        if (p.val < root.val) {
            succ = root;
            root = root.left;
        } else {
            root = root.right;
        }
    }
    return succ;
}

What is the main bug in this code?

medium

A. It uses wrong comparison operator in while loop

B. It incorrectly updates successor when p.val >= root.val

C. It does not handle the case when p has no right child

D. It may return null even if successor exists

Solution

1. Step 1: Analyze right subtree handling

   If p has right child, code correctly finds leftmost node in right subtree.

2. Step 2: Analyze ancestor search loop

   Loop updates successor when p.val < root.val, else moves right. This is correct logic.

3. Step 3: Identify potential null return

   If p is the largest node, no successor exists, so null return is valid. Otherwise, code works.

4. Final Answer:

   It may return null even if successor exists -> Option D

5. Quick Check:

   Null return possible if logic misses successor [OK]

Hint: Check if successor can be missed when traversing ancestors [OK]

Common Mistakes:

• Assuming null means no successor always
• Ignoring right subtree case
• Misreading comparison operators

5. You have a BST and want to find the inorder successor of node with value 25. The BST is:

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

Which node is the inorder successor of 25 and why?

hard

A. 30, because it is the smallest node greater than 25 in the right subtree

B. 50, because it is the leftmost node in the right subtree of 25

C. 40, because 25 has no right child and 40 is the nearest bigger ancestor

D. 20, because it is the parent of 25

Solution

1. Step 1: Check if node 25 has right child

   Node 25 has no right child, so successor is ancestor or next bigger node.

2. Step 2: Find nearest ancestor greater than 25

   Node 40 is the nearest ancestor greater than 25, so it is the inorder successor.

3. Final Answer:

   40, because 25 has no right child and 40 is the nearest bigger ancestor -> Option C

4. Quick Check:

   Next bigger node after 25 is 40 [OK]

Hint: Successor is leftmost node in right subtree or nearest bigger ancestor [OK]

Common Mistakes:

• Choosing 30 which is smaller than 25
• Choosing 50 which is too far
• Choosing parent 20 which is smaller