Two Sum in BST in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the BST node is null.

DSA Javascript

if (root [1] null) return false;

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

A!==

B===

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of ===

Using assignment = instead of comparison ===

We check if root is exactly equal to null to stop recursion.

2fill in blank

medium

Complete the code to add a value to the set.

DSA Javascript

set.[1](root.val);

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

Apush

Binsert

Cappend

Dadd

Attempts:

3 left

💡 Hint

Common Mistakes

Using push() which is for arrays

Using insert() which is not a Set method

JavaScript Set uses add() to insert values.

3fill in blank

hard

Fix the error in checking if the complement exists in the set.

DSA Javascript

if (set.[1](target - root.val)) return true;

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

Ahas

Bincludes

Cfind

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Using contains() which is not a Set method

Using includes() which is for arrays

Set uses has() method to check for presence of an element.

4fill in blank

hard

Fill both blanks to complete the recursive calls for left and right subtrees.

DSA Javascript

return dfs(root.[1], set, target) || dfs(root.[2], set, target);

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

Aleft

Bright

Cchild

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect property names like child or node

Calling dfs on the same child twice

We recursively call dfs on left and right children of the node.

5fill in blank

hard

Fill all three blanks to complete the main function that uses DFS and a set.

DSA Javascript

var findTarget = function(root, target) {
  const set = new Set();
  function dfs(node, set, target) {
    if (node [1] null) return false;
    if (set.[2](target - node.val)) return true;
    set.[3](node.val);
    return dfs(node.left, set, target) || dfs(node.right, set, target);
  }
  return dfs(root, set, target);
};

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

A===

Bhas

Cadd

D!==

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of === for null check

Using contains instead of has for set

Using push instead of add for set

Check if node is null with ===, check set with has(), add node.val with add().

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