Two Sum in BST in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Two Sum problem in a Binary Search Tree (BST)?

To find if there exist two nodes in the BST such that their values add up to a given target sum.

intermediate

Why is BST property useful in solving the Two Sum problem efficiently?

Because BST nodes are ordered, we can use two pointers (one from smallest, one from largest) to find pairs that sum to the target without checking all pairs.

beginner

What data structure can be used to store visited node values during a DFS to solve Two Sum in BST?

A HashSet can be used to store visited node values to quickly check if the complement (target - current node value) exists.

advanced

Explain the two-pointer approach for Two Sum in BST.

Use an iterator to get the next smallest and another for the next largest node. Move pointers inward based on sum comparison until a pair is found or pointers cross.

intermediate

What is the time complexity of the Two Sum in BST solution using HashSet and DFS?

O(n), where n is the number of nodes, because each node is visited once and HashSet lookups are O(1) on average.

What data structure helps check complements quickly in Two Sum in BST?

A HashSet

B Queue

C Stack

D ArrayList

HashSet allows O(1) average time lookup to check if complement exists.

Which traversal is commonly used to get sorted node values from a BST?

A Inorder traversal

B Preorder traversal

C Postorder traversal

D Level order traversal

Inorder traversal visits BST nodes in ascending order.

In the two-pointer approach, what do the pointers represent?

A Root and leaf nodes

B Left and right child nodes

C Smallest and largest node values

D Parent and child nodes

Pointers start at smallest and largest values to find pairs summing to target.

What is the worst-case time complexity of Two Sum in BST using two-pointer approach?

A O(log n)

B O(n)

C O(n log n)

D O(n^2)

Each node is visited at most once, so time complexity is linear.

Which of these is NOT a valid approach to solve Two Sum in BST?

A Using HashSet during DFS

B Using two iterators for smallest and largest values

C Sorting all node values and using binary search for complements

D Using BFS without any additional data structure

BFS alone without extra data structures does not efficiently solve Two Sum in BST.

Describe how you would use a HashSet and DFS to solve the Two Sum problem in a BST.

Think about storing visited values and checking complements.

Explain the two-pointer technique to find two nodes in a BST that sum to a target value.

Imagine walking from both ends of a sorted list.

Practice

1. What is the main goal of the Two Sum in BST problem?

easy

A. Count the total number of nodes in the BST.

B. Find the maximum value in the BST.

C. Find two nodes in the BST whose values add up to a target sum.

D. Check if the BST is balanced.

Solution

1. Step 1: Understand the problem statement

   The problem asks to find if there exist two nodes in the BST such that their values add up to a given target.

2. Step 2: Identify the goal

   The goal is to check for the existence of such a pair, not to find max, balance, or count nodes.

3. Final Answer:

   Find two nodes in the BST whose values add up to a target sum. -> Option C

4. Quick Check:

   Two Sum in BST means finding two nodes summing to target [OK]

Hint: Remember Two Sum means pair sums to target [OK]

Common Mistakes:

• Confusing Two Sum with finding max value
• Thinking problem is about BST balance
• Counting nodes instead of finding pairs

2. Which of the following Java method signatures correctly represents a function to solve Two Sum in BST?

easy

A. public int findSum(TreeNode root, int k)

B. public boolean findTarget(TreeNode root, int k)

C. public void findTarget(TreeNode root, int k)

D. public TreeNode findPair(TreeNode root, int k)

Solution

1. Step 1: Understand the return type needed

   The function should return true or false if a pair exists, so return type is boolean.

2. Step 2: Check method signatures

   public boolean findTarget(TreeNode root, int k) returns boolean and accepts root and target, matching the problem requirements.

3. Final Answer:

   public boolean findTarget(TreeNode root, int k) -> Option B

4. Quick Check:

   Boolean return for existence check [OK]

Hint: Two Sum returns true/false, so use boolean return [OK]

Common Mistakes:

• Using void return type instead of boolean
• Returning int instead of boolean
• Returning TreeNode instead of boolean

3. Given the BST below and target = 9, what will the output of findTarget(root, 9) be?

      5
     / \
    3   6
   / \   \
  2   4   7

medium

A. true

B. false

C. null

D. Runtime Error

Solution

1. Step 1: Identify pairs that sum to 9

   Check pairs: 2+7=9, 3+6=9, 4+5=9. All valid pairs exist.

2. Step 2: Confirm function returns true if any pair found

   Since pairs exist, function returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Pairs 2+7 or 3+6 sum to 9 [OK]

Hint: Check pairs adding to target in BST nodes [OK]

Common Mistakes:

• Ignoring valid pairs and returning false
• Confusing output with node values
• Assuming null or error without reason

4. Consider this buggy code snippet for Two Sum in BST:

public boolean findTarget(TreeNode root, int k) {
    Set<Integer> set = new HashSet<>();
    return dfs(root, k, set);
}

private boolean dfs(TreeNode node, int k, Set<Integer> set) {
    if (node == null) return false;
    if (set.contains(k - node.val)) return true;
    set.add(node.val);
    dfs(node.left, k, set);
    dfs(node.right, k, set);
    return false;
}

What is the main bug causing incorrect results?

medium

A. Using HashSet instead of ArrayList.

B. Adding node values after checking set.

C. Not checking if root is null in findTarget.

D. Missing return statements from recursive calls in dfs.

Solution

1. Step 1: Analyze dfs recursive calls

   dfs calls itself on left and right but ignores their return values.

2. Step 2: Understand impact of missing returns

   Without returning results of recursive calls, function always returns false unless first node matches.

3. Step 3: Correct fix

   Return true if either left or right dfs call returns true.

4. Final Answer:

   Missing return statements from recursive calls in dfs. -> Option D

5. Quick Check:

   Return recursive calls to propagate true [OK]

Hint: Always return recursive call results in DFS [OK]

Common Mistakes:

• Ignoring return values of recursive calls
• Using wrong data structure for lookup
• Not handling null root properly

5. You want to optimize Two Sum in BST to use O(h) space, where h is tree height. Which approach is best?

hard

A. Use two iterators: one in-order and one reverse in-order traversal.

B. Store all node values in a list and use two pointers.

C. Use a HashSet to store visited nodes during DFS.

D. Convert BST to sorted array and binary search for complements.

Solution

1. Step 1: Understand space complexity of approaches

   Storing all nodes uses O(n) space; HashSet also O(n); array conversion O(n).

2. Step 2: Two iterator approach

   Using two iterators (in-order and reverse in-order) simulates two pointers with O(h) space.

3. Step 3: Choose best approach

   Use two iterators: one in-order and one reverse in-order traversal. uses O(h) space and is optimal for BST Two Sum.

4. Final Answer:

   Use two iterators: one in-order and one reverse in-order traversal. -> Option A

5. Quick Check:

   Two iterators give O(h) space [OK]

Hint: Two pointers in BST use two iterators for O(h) space [OK]

Common Mistakes:

• Using full list storage causing O(n) space
• Relying on HashSet which uses O(n) space
• Ignoring BST properties for optimization