Two Sum in BST in DSA Java - Time & Space Complexity

Time Complexity: Two Sum in BST

O(n)

Understanding Time Complexity

We want to know how the time needed to find two numbers that add up to a target grows as the tree gets bigger.

How does the search time change when the number of nodes in the BST increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

This code searches the BST to find if any two nodes sum to the target k using a depth-first search and a set to track visited values.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once in the tree using recursion.
• How many times: Once per node, so total visits equal the number of nodes n.

How Execution Grows With Input

As the number of nodes grows, the function checks each node once, so the work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits and checks
100	About 100 visits and checks
1000	About 1000 visits and checks

Pattern observation: The operations grow linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the two sum grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Because the tree is a BST, searching for pairs is faster than visiting all nodes."

[OK] Correct: The code visits every node once to check pairs, so it still needs to look at all nodes in the worst case.

Interview Connect

Understanding how to analyze tree traversal time helps you explain your approach clearly and shows you know how your solution scales with input size.

Self-Check

"What if we used an in-order traversal to get a sorted list first, then used two pointers to find the sum? How would the time complexity change?"

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