Two Sum in BST in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

if root [1] nil {
    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 '==' causes the logic to be reversed.

Using comparison operators like '<' or '>' with nil causes errors.

We check if the root is equal to nil to know if we reached the end of a branch.

2fill in blank

medium

Complete the code to insert a value into the map to track visited numbers.

DSA Go

visited[[1]] = true

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

Asum

Btarget

Ccomplement

Droot.Val

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'target' or 'sum' instead of the node's value.

Using 'complement' which is the needed pair value, not the current node.

We mark the current node's value as visited by setting visited[root.Val] = true.

3fill in blank

hard

Fix the error in the condition to check if the complement exists in the map.

DSA Go

if _, ok := visited[[1]]; ok {
    return true
}

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

Atarget - root.Val

Broot.Val - target

Croot.Val + target

Dtarget + root.Val

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of subtraction.

Reversing the subtraction order.

The complement is target minus the current node's value to find the needed pair.

4fill in blank

hard

Fill both blanks to recursively search left and right subtrees.

DSA Go

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

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

ALeft

BRight

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using uppercase 'Left' or 'Right' causes compilation errors.

Swapping left and right fields.

In Go, struct fields start with lowercase letters for unexported fields, so use 'left' and 'right'.

5fill in blank

hard

Fill all three blanks to complete the TwoSumBST function using DFS and a map.

DSA Go

func TwoSumBST(root *TreeNode, target int) bool {
    visited := make(map[int]bool)
    var dfs func(node *TreeNode, target int, visited map[int]bool) bool
    dfs = func(node *TreeNode, target int, visited map[int]bool) bool {
        if node [1] nil {
            return false
        }
        complement := target [2] node.Val
        if _, ok := visited[complement]; ok {
            return true
        }
        visited[node.Val] = true
        return dfs(node.[3], target, visited) || dfs(node.right, target, visited)
    }
    return dfs(root, target, visited)
}

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

A==

B-

Cleft

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' for nil check.

Adding instead of subtracting for complement.

Using uppercase 'Left' instead of 'left'.

Check if node is nil with '==', find complement by subtracting node.Val from target, and recurse left subtree with 'left'.

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