Two Sum in BST in DSA Typescript - 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 increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function findTarget(root: TreeNode | null, k: number): boolean {
  const set = new Set();
  function dfs(node: TreeNode | null): boolean {
    if (!node) return false;
    if (set.has(k - node.val)) return true;
    set.add(node.val);
    return dfs(node.left) || dfs(node.right);
  }
  return dfs(root);
}
    

This code searches the BST to find if two values sum to k by traversing all nodes once and checking complements in a set.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive traversal of each node in the BST.
• How many times: Each node is visited exactly once during the depth-first search.

How Execution Grows With Input

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

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

Pattern observation: The operations increase 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 takes time proportional to all nodes, not less.

Interview Connect

Understanding this helps you explain how traversing a tree and using extra space for quick lookups works together, a common pattern in coding problems.

Self-Check

"What if we used an in-order traversal to create a sorted array 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. Check if a BST is balanced

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

C. Count the number of nodes in a BST

D. Find the maximum value in a BST

Solution

1. Step 1: Understand the problem statement

   The problem asks to find two nodes in a Binary Search Tree (BST) such that their values add up to a given target sum.

2. Step 2: Identify the goal

   The goal is to check if such a pair exists, not to find maximum, count nodes, or check balance.

3. Final Answer:

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

4. Quick Check:

   Two Sum in BST = Find pair with target sum [OK]

Hint: Look for two values adding to target in BST [OK]

Common Mistakes:

• Confusing with finding max or min value
• Thinking it counts nodes instead of finding pairs
• Assuming it checks tree balance

2. Which TypeScript syntax correctly defines a BST node for the Two Sum problem?

easy

A. interface TreeNode { val: string; left: TreeNode; right: TreeNode; }

B. function TreeNode(val) { this.val = val; this.left = null; this.right = null; }

C. type TreeNode = { val: number; left?: TreeNode; right?: TreeNode; }

D. class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } }

Solution

1. Step 1: Check correct TypeScript class syntax for BST node

   class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } } defines a class with number value and nullable left/right children, matching BST node structure.

2. Step 2: Compare other options

   function TreeNode(val) { this.val = val; this.left = null; this.right = null; } is JavaScript function style, not TypeScript class. interface TreeNode { val: string; left: TreeNode; right: TreeNode; } uses string type for val and non-nullable children, which is incorrect. type TreeNode = { val: number; left?: TreeNode; right?: TreeNode; } uses optional properties but lacks constructor and proper null handling.

3. Final Answer:

   class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } } -> Option D

4. Quick Check:

   Correct BST node class syntax = class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } } [OK]

Hint: Use class with nullable left/right and number val [OK]

Common Mistakes:

• Using string type for node value
• Not initializing left and right to null
• Using JavaScript function instead of TypeScript class

3. Given the BST below and target = 9, what will the Two Sum function return?

    5
   / \
  3   7
 /     \
2       8

Function uses in-order traversal and a set to find pairs.

medium

A. true

B. false

C. null

D. Error

Solution

1. Step 1: Traverse BST in-order and check pairs

   In-order traversal gives sorted values: [2, 3, 5, 7, 8]. We check if target - current value exists in set.

2. Step 2: Find pair summing to 9

   Check pairs: 2 + 7 = 9 (found), so function returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   2 + 7 = 9 found in BST [OK]

Hint: In-order traversal gives sorted list to find pairs [OK]

Common Mistakes:

• Assuming no pair without checking all nodes
• Confusing traversal order
• Ignoring set lookup for complements

4. Identify the bug in this TypeScript Two Sum in BST function snippet:

function findTarget(root: TreeNode | null, k: number): boolean {
  const set = new Set<number>();
  function dfs(node: TreeNode | null): boolean {
    if (!node) return false;
    if (set.has(k - node.val)) return true;
    set.add(node.val);
    dfs(node.left);
    dfs(node.right);
    return false;
  }
  return dfs(root);
}

medium

A. The base case should return true when node is null

B. Set is not initialized properly

C. The function returns false too early, missing true from recursive calls

D. The function should use a queue instead of recursion

Solution

1. Step 1: Analyze recursive calls and return values

   The dfs function calls dfs(node.left) and dfs(node.right) but ignores their return values, always returning false at the end.

2. Step 2: Identify missing return propagation

   If dfs(node.left) or dfs(node.right) returns true, dfs should return true immediately to propagate the success.

3. Final Answer:

   The function returns false too early, missing true from recursive calls -> Option C

4. Quick Check:

   Return recursive true results to avoid missing pairs [OK]

Hint: Return recursive calls' results to propagate true [OK]

Common Mistakes:

• Ignoring return values of recursive calls
• Returning false before checking all nodes
• Misunderstanding base case return

5. You want to optimize Two Sum in BST for large trees. Which approach best reduces time complexity while using O(h) space, where h is tree height?

hard

A. Use two pointers with iterative in-order and reverse in-order traversals simultaneously

B. Convert BST to array and use nested loops to find pairs

C. Use breadth-first search with a hash set to find pairs

D. Perform recursive DFS and store all node values in a list before checking pairs

Solution

1. Step 1: Understand space and time constraints

   We want to reduce time complexity and keep space to O(h), where h is tree height.

2. Step 2: Evaluate approaches

   Use two pointers with iterative in-order and reverse in-order traversals simultaneously uses two pointers with iterative in-order (smallest to largest) and reverse in-order (largest to smallest) traversals, allowing pair search in O(n) time and O(h) space. Other options use extra space O(n) or nested loops with O(n^2) time.

3. Final Answer:

   Use two pointers with iterative in-order and reverse in-order traversals simultaneously -> Option A

4. Quick Check:

   Two pointer BST traversal = O(n) time, O(h) space [OK]

Hint: Use two pointers with in-order and reverse in-order iterators [OK]

Common Mistakes:

• Using nested loops causing O(n^2) time
• Storing all nodes in list increasing space to O(n)
• Using BFS which uses more space than O(h)