Two Sum in BST in DSA Javascript - 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 whose values add up to a given target sum.

intermediate

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

Because BST nodes are ordered, we can use in-order traversal to get sorted values, which helps in efficiently finding pairs that sum to the target.

intermediate

What is a common approach to solve Two Sum in BST using extra space?

Perform an in-order traversal to get a sorted array of node values, then use two pointers (start and end) to find if any two values sum to the target.

advanced

How can we solve Two Sum in BST without converting it to an array?

Use two iterators: one for the smallest value (in-order) and one for the largest value (reverse in-order), then move them towards each other to find the target sum.

intermediate

What is the time complexity of the Two Sum in BST solution using in-order traversal and two pointers?

O(n), where n is the number of nodes in the BST, because we visit each node once during traversal and then use two pointers to find the pair.

What traversal method is commonly used to get sorted values from a BST?

A Level-order traversal

B In-order traversal

C Post-order traversal

D Pre-order traversal

In-order traversal visits BST nodes in ascending order, giving sorted values.

In the Two Sum problem on BST, what data structure helps to check if a complement value exists quickly?

A Stack

B Queue

C Hash set

D Linked list

A hash set allows O(1) average time lookup to check if the complement of the current node's value exists.

Which technique uses two pointers to find two numbers that add up to a target in a sorted array?

A Two-pointer technique

B Binary search

C Depth-first search

D Breadth-first search

Two-pointer technique moves pointers from start and end towards each other to find the target sum.

What is the space complexity of the Two Sum in BST solution using in-order traversal and array conversion?

A O(n)

B O(1)

C O(log n)

D O(n log n)

Storing all node values in an array requires O(n) space.

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

A Using two iterators from smallest and largest nodes

B Using in-order traversal and two pointers on array

C Using a hash set to store visited values

D Using post-order traversal only without extra data structures

Post-order traversal alone does not help efficiently find pairs without extra data structures.

Explain how to solve the Two Sum problem in a BST using in-order traversal and two pointers.

Think about how sorted arrays help find pairs with a target sum.

Describe the space-optimized approach to find two nodes in a BST that sum to a target without converting the BST to an array.

Imagine walking from smallest and largest values at the same time.

Practice

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

easy

A. Find the maximum value in the BST.

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

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

D. Check if the BST is balanced.

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 max value, count nodes, or check balance.

3. Final Answer:

   Find two nodes in the 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 pairs adding to target in BST nodes [OK]

Common Mistakes:

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

2. Which of the following is the correct way to declare a BST node in JavaScript for the Two Sum problem?

easy

A. function Node(val) { this.val = val; this.left = null; this.right = null; }

B. let Node = (val) => { val = val; left = null; right = null; }

C. class Node { constructor(val) { val = val; left = null; right = null; } }

D. var Node = function(val) { this.value = val; this.left = this.right = undefined; }

Solution

1. Step 1: Check correct property assignments

   function Node(val) { this.val = val; this.left = null; this.right = null; } correctly assigns this.val, this.left, and this.right to initialize a BST node.

2. Step 2: Identify syntax and property errors in other options

   let Node = (val) => { val = val; left = null; right = null; } misses this keyword and uses arrow function incorrectly. class Node { constructor(val) { val = val; left = null; right = null; } } assigns to local variables, not this. var Node = function(val) { this.value = val; this.left = this.right = undefined; } uses this.value instead of this.val and sets children to undefined instead of null.

3. Final Answer:

   function Node(val) { this.val = val; this.left = null; this.right = null; } -> Option A

4. Quick Check:

   Correct node declaration uses this.val and null children [OK]

Hint: Use function with this.val, this.left, this.right = null [OK]

Common Mistakes:

• Forgetting 'this' keyword for properties
• Using arrow functions for constructor
• Assigning children to undefined instead of null

3. Given the BST below and target = 9, what will the function findTarget(root, 9) return?

    5
   / \
  3   6
 / \   \
2   4   7

Assuming findTarget returns true if two nodes sum to target, else false.

medium

A. true

B. false

C. null

D. undefined

Solution

1. Step 1: Identify pairs that sum to 9

   Check pairs: 2+7=9, 3+6=9, 4+5=9. All these pairs exist in the BST.

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

   Since pairs exist, findTarget returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Pairs sum to 9 = true [OK]

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

Common Mistakes:

• Ignoring valid pairs in BST
• Assuming function returns node instead of boolean
• Confusing null or undefined as return

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

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

What is the main bug causing incorrect results?

medium

A. Using Set instead of Array for storage

B. Checking for k - node.val before adding node.val

C. Missing return statements in recursive calls inside dfs

D. Not handling empty tree case

Solution

1. Step 1: Analyze recursive calls in dfs

   The recursive calls to dfs(node.left) and dfs(node.right) do not return their results, so true found deeper is ignored.

2. Step 2: Fix by returning recursive calls

   Adding return before recursive calls ensures if true is found, it propagates up and final result is correct.

3. Final Answer:

   Missing return statements in recursive calls inside dfs -> Option C

4. Quick Check:

   Return recursive calls to propagate true [OK]

Hint: Always return recursive calls to propagate results [OK]

Common Mistakes:

• Not returning recursive call results
• Confusing Set usage as bug
• Ignoring base case handling

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

hard

A. Convert BST to a linked list and then find pairs.

B. Store all node values in an array and use a hash set to find pairs.

C. Perform a breadth-first search and check pairs at each level.

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

Solution

1. Step 1: Understand space complexity of approaches

   Storing all nodes uses O(n) space. BFS also uses O(n) space. Converting to linked list uses O(n).

2. Step 2: Use two iterators approach

   Using two iterators (one in-order for smallest, one reverse in-order for largest) uses O(h) space, where h is tree height, optimizing space.

3. Final Answer:

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

4. Quick Check:

   Two iterators = O(h) space optimization [OK]

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

Common Mistakes:

• Using full array storage increasing space to O(n)
• Trying BFS which uses more space
• Converting BST to linked list wastes space