Two Sum in BST in DSA Go - 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?

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.

beginner

What data structure is commonly used to check for complements when solving Two Sum in BST?

A hash set is 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.

Perform in-order traversal to get sorted values, then use two pointers at start and end to find if any two values sum to the target by moving pointers inward based on sum comparison.

intermediate

What is the time complexity of solving Two Sum in BST using a hash set during traversal?

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

What traversal method is best to get sorted node values from a BST?

A In-order traversal

B Pre-order traversal

C Post-order traversal

D Level-order traversal

In-order traversal visits nodes in ascending order for BSTs.

Which data structure helps to check if a complement value exists during Two Sum in BST?

A Stack

B Linked list

C Queue

D Hash set

Hash set allows O(1) average time lookup for complements.

If you use two pointers on a sorted array of BST values, what do you do if the sum is less than the target?

A Move the right pointer left

B Move the left pointer right

C Stop searching

D Move both pointers

Moving the left pointer right increases the sum.

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

A O(n)

B O(n log n)

C O(log n)

D O(n^2)

Each node is visited once, so O(n).

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

A In-order traversal + hash set

B In-order traversal + two pointers

C Pre-order traversal + two pointers

D Level-order traversal + hash set

Pre-order traversal does not produce sorted values, so it is not suitable for two-pointer approach but can be used with hash set. The question is about valid approaches; pre-order + hash set is valid but less common.

Describe how you would find two nodes in a BST that sum to a target value using a hash set.

Think about storing visited values and checking complements.

Explain the two-pointer technique to solve Two Sum in BST and why it requires in-order traversal.

Sorting is key to two-pointer method.

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 given 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, count nodes, or check balance.

3. Final Answer:

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

4. Quick Check:

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

Hint: Look for two values that add up to target in BST [OK]

Common Mistakes:

• Confusing with finding max or min value
• Trying to count nodes instead of finding pairs
• Checking BST balance instead of sum pairs

2. Which of the following Go function signatures is correct for a function that checks if two sum exists in a BST?

easy

A. func findTarget(root *TreeNode, k int) bool

B. func findTarget(root TreeNode, k string) int

C. func findTarget(root *TreeNode, k float64) string

D. func findTarget(root TreeNode, k bool) *TreeNode

Solution

1. Step 1: Understand function parameters and return type

   The function should take a pointer to the root node of BST and an integer target sum, returning a boolean indicating if pair exists.

2. Step 2: Match correct signature

   func findTarget(root *TreeNode, k int) bool matches: pointer to TreeNode, integer target, returns bool.

3. Final Answer:

   func findTarget(root *TreeNode, k int) bool -> Option A

4. Quick Check:

   Correct function signature = func findTarget(root *TreeNode, k int) bool [OK]

Hint: Function takes *TreeNode and int, returns bool [OK]

Common Mistakes:

• Using wrong parameter types like string or bool
• Passing TreeNode by value instead of pointer
• Returning wrong type like int or string

3. What will be the output of the following Go code snippet for Two Sum in BST?

type TreeNode struct {
    Val int
    Left, Right *TreeNode
}

func findTarget(root *TreeNode, k int) bool {
    seen := make(map[int]bool)
    var dfs func(*TreeNode) bool
    dfs = func(node *TreeNode) bool {
        if node == nil {
            return false
        }
        if seen[k - node.Val] {
            return true
        }
        seen[node.Val] = true
        return dfs(node.Left) || dfs(node.Right)
    }
    return dfs(root)
}

root := &TreeNode{Val:5, Left:&TreeNode{Val:3}, Right:&TreeNode{Val:6}}
result := findTarget(root, 9)
fmt.Println(result)

medium

A. true

B. false

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Trace the DFS traversal and map usage

   The function uses DFS to traverse nodes. It checks if (target - current node value) is in the map. For root 5, seen is empty, adds 5. Goes left to 3, checks if 9-3=6 in seen (no), adds 3. Goes right to 6, checks if 9-6=3 in seen (yes), returns true.

2. Step 2: Confirm output

   Since pair (3,6) sums to 9, function returns true and prints true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Pair found (3+6=9) = true [OK]

Hint: Check if complement exists in map during DFS [OK]

Common Mistakes:

• Not checking complement before adding current value
• Returning false too early
• Ignoring nil node checks causing panic

4. Identify the bug in this Go code for Two Sum in BST and choose the fix:

func findTarget(root *TreeNode, k int) bool {
    seen := make(map[int]bool)
    var dfs func(*TreeNode) bool
    dfs = func(node *TreeNode) bool {
        if node == nil {
            return false
        }
        seen[node.Val] = true
        if seen[k - node.Val] {
            return true
        }
        return dfs(node.Left) || dfs(node.Right)
    }
    return dfs(root)
}

medium

A. Add a nil check after adding to map.

B. Remove the map and use a slice instead.

C. Change return type to int instead of bool.

D. Check complement before adding current node value to map.

Solution

1. Step 1: Analyze the order of operations in DFS

   The code adds current node value to map before checking if complement exists. This causes the complement check to always find current value, leading to false positives.

2. Step 2: Correct the order

   Check if complement exists in map before adding current node value to avoid matching the same node twice.

3. Final Answer:

   Check complement before adding current node value to map. -> Option D

4. Quick Check:

   Complement check before map insert fixes bug [OK]

Hint: Check complement before adding current value [OK]

Common Mistakes:

• Adding current value before checking complement
• Using wrong data structure for lookup
• Ignoring nil node checks

5. Given a BST with nodes [4, 2, 6, 1, 3, 5, 7], which pairs sum to 9? Select the correct output of a Two Sum in BST function that returns all pairs.

func findAllPairs(root *TreeNode, k int) [][2]int {
    seen := make(map[int]bool)
    pairs := [][2]int{}
    var dfs func(*TreeNode)
    dfs = func(node *TreeNode) {
        if node == nil {
            return
        }
        complement := k - node.Val
        if seen[complement] {
            pairs = append(pairs, [2]int{complement, node.Val})
        }
        seen[node.Val] = true
        dfs(node.Left)
        dfs(node.Right)
    }
    dfs(root)
    return pairs
}

hard

A. [[5,4],[6,3],[7,2]]

B. [[1,8],[2,7],[3,6]]

C. [[3,6],[4,5],[2,7]]

D. [[1,7],[2,6],[3,5]]

Solution

1. Step 1: Identify all pairs in BST that sum to 9

   Nodes are 1,2,3,4,5,6,7. Pairs summing to 9 are (2,7), (3,6), (4,5).

2. Step 2: Match pairs with function output format

   The function appends pairs as [complement, node.Val] during preorder traversal: at 6 finds 3 in seen, appends [3,6]; at 5 finds 4, [4,5]; at 7 finds 2, [2,7].

3. Final Answer:

   [[3,6],[4,5],[2,7]] -> Option C

4. Quick Check:

   Pairs summing to 9 in traversal order = [[3,6],[4,5],[2,7]] [OK]

Hint: List pairs where values sum to target k [OK]

Common Mistakes:

• Including pairs with values not in BST
• Reversing pair order incorrectly
• Missing valid pairs