Top View of Binary Tree in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new queue for BFS traversal.

DSA Go

queue := make(chan *Node, [1])

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

A0

B1

C100

D10

Attempts:

3 left

💡 Hint

Common Mistakes

Using zero capacity channel causes deadlock.

Using capacity 1 is too small for BFS queue.

We create a buffered channel with capacity 10 to hold nodes during BFS.

2fill in blank

medium

Complete the code to update the horizontal distance map only if the distance is not already present.

DSA Go

if _, ok := hdMap[hd]; ![1] { hdMap[hd] = node.data }

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

Aok

Bexists

Cfound

Dpresent

Attempts:

3 left

💡 Hint

Common Mistakes

Using variable names other than 'ok' for map presence check.

Not negating the condition to insert only once.

The idiomatic way in Go to check map presence is using 'ok'.

3fill in blank

hard

Fix the error in the BFS loop to correctly update horizontal distance and enqueue children.

DSA Go

for len(queue) > 0 {
    node := <-queue
    hd := [1][node]
    if _, ok := hdMap[hd]; !ok {
        hdMap[hd] = node.data
    }
    if node.left != nil {
        queue <- node.left
        [2][node.left] = hd - 1
    }
    if node.right != nil {
        queue <- node.right
        hdDist[node.right] = hd + 1
    }
}

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

AhdMap

BhdDist

Cdistance

Dhd

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'hdMap' to store horizontal distances causes wrong data.

Not updating horizontal distances for children.

We use 'hdDist' map to track horizontal distances of nodes, not 'hdMap' which stores data.

4fill in blank

hard

Fill both blanks to correctly initialize the horizontal distance map and enqueue the root node.

DSA Go

hdDist := make(map[*Node]int)
hdDist[[1]] = 0
queue <- [2]

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

Aroot

Bhead

CrootNode

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for root causes errors.

Not initializing horizontal distance for root.

We set the root node's horizontal distance to 0 and enqueue the root node.

5fill in blank

hard

Fill all three blanks to print the top view nodes in order of horizontal distance.

DSA Go

keys := make([]int, 0, len(hdMap))
for k := range hdMap {
    keys = append(keys, k)
}
sort.[1](sort.IntSlice(keys))
for _, k := range keys {
    fmt.[2](hdMap[k], " ")
}
fmt.[3]()

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

ASortInts

BPrintln

CPrint

DSort

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong sort function causes compile error.

Using fmt.Println inside loop prints each node on new line.

We use sort.Sort with sort.IntSlice, print with fmt.Print, and end line with fmt.Println.

Practice

1. What does the top view of a binary tree represent?

easy

A. All leaf nodes of the tree

B. Nodes visible when the tree is viewed from the left side

C. Nodes visible when the tree is viewed from above

D. Nodes visible when the tree is viewed from the right side

Solution

1. Step 1: Understand the top view concept

   The top view of a binary tree includes all nodes visible when looking down vertically from above the tree.

2. Step 2: Differentiate from other views

   Left or right views show nodes from side angles, while top view shows nodes from above.

3. Final Answer:

   Nodes visible when the tree is viewed from above -> Option C

4. Quick Check:

   Top view = nodes seen from above [OK]

Hint: Top view shows nodes visible from above only [OK]

Common Mistakes:

• Confusing top view with left or right view
• Including all leaf nodes as top view
• Thinking top view shows all nodes

2. Which Go data structure is best suited to store nodes for computing the top view of a binary tree?

easy

A. Map with horizontal distance as key

B. Array

C. Slice

D. Linked list

Solution

1. Step 1: Identify the need for horizontal distance mapping

   Top view requires tracking nodes by their horizontal distance from root to avoid duplicates.

2. Step 2: Choose data structure supporting key-value pairs

   A map with horizontal distance as key stores the first node at each horizontal distance efficiently.

3. Final Answer:

   Map with horizontal distance as key -> Option A

4. Quick Check:

   Use map for horizontal distance keys [OK]

Hint: Use map keyed by horizontal distance for top view [OK]

Common Mistakes:

• Using slice or array without horizontal distance keys
• Using linked list which lacks key access
• Ignoring horizontal distance concept

3. Given the following Go code snippet for top view, what will be printed?

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

func topView(root *Node) []int {
  if root == nil {
    return []int{}
  }
  type pair struct { node *Node; hd int }
  queue := []pair{{root, 0}}
  m := map[int]int{}
  res := []int{}

  for len(queue) > 0 {
    p := queue[0]
    queue = queue[1:]
    if _, ok := m[p.hd]; !ok {
      m[p.hd] = p.node.Val
    }
    if p.node.Left != nil {
      queue = append(queue, pair{p.node.Left, p.hd - 1})
    }
    if p.node.Right != nil {
      queue = append(queue, pair{p.node.Right, p.hd + 1})
    }
  }

  keys := []int{}
  for k := range m {
    keys = append(keys, k)
  }
  sort.Ints(keys)
  for _, k := range keys {
    res = append(res, m[k])
  }
  return res
}

root := &Node{Val: 1}
root.Left = &Node{Val: 2}
root.Right = &Node{Val: 3}
root.Left.Right = &Node{Val: 4}
root.Left.Right.Right = &Node{Val: 5}
root.Left.Right.Right.Right = &Node{Val: 6}

fmt.Println(topView(root))

medium

A. [2 1 3]

B. [2 1 3 6]

C. [2 1 4 5 6]

D. [1 2 3 4 5 6]

Solution

1. Step 1: Traverse nodes with horizontal distances

   Root(1) at hd=0, Left(2) at hd=-1, Right(3) at hd=1, Left.Right(4) at hd=0 (ignored), Left.Right.Right(5) at hd=1 (ignored), Left.Right.Right.Right(6) at hd=2.

2. Step 2: Collect first nodes at each horizontal distance

   hd=-1:2, hd=0:1, hd=1:3, hd=2:6

3. Final Answer:

   [2 1 3 6] -> Option B

4. Quick Check:

   Top view nodes by hd sorted: 2,1,3,6 [OK]

Hint: First node at each horizontal distance forms top view [OK]

Common Mistakes:

• Including nodes overwritten at same horizontal distance
• Ignoring horizontal distance ordering
• Confusing top view with level order

4. The following Go code attempts to compute the top view of a binary tree but has a bug. What is the bug?

func topView(root *Node) []int {
  if root == nil {
    return []int{}
  }
  type pair struct { node *Node; hd int }
  queue := []pair{{root, 0}}
  m := map[int]int{}
  res := []int{}

  for len(queue) > 0 {
    p := queue[0]
    queue = queue[1:]
    m[p.hd] = p.node.Val
    if p.node.Left != nil {
      queue = append(queue, pair{p.node.Left, p.hd - 1})
    }
    if p.node.Right != nil {
      queue = append(queue, pair{p.node.Right, p.hd + 1})
    }
  }

  keys := []int{}
  for k := range m {
    keys = append(keys, k)
  }
  sort.Ints(keys)
  for _, k := range keys {
    res = append(res, m[k])
  }
  return res
}

medium

A. It overwrites nodes at the same horizontal distance, losing top view nodes

B. It does not handle nil root properly

C. It uses a slice instead of a queue

D. It does not sort the keys before output

Solution

1. Step 1: Analyze map assignment inside loop

   The code assigns m[p.hd] = p.node.Val every time, overwriting previous values at same horizontal distance.

2. Step 2: Understand top view requirement

   Top view requires storing only the first node encountered at each horizontal distance, so overwriting causes incorrect output.

3. Final Answer:

   It overwrites nodes at the same horizontal distance, losing top view nodes -> Option A

4. Quick Check:

   Overwrite causes wrong top view [OK]

Hint: Store first node per horizontal distance only [OK]

Common Mistakes:

• Overwriting map entries instead of checking existence
• Ignoring horizontal distance uniqueness
• Not sorting keys before output

5. You have a binary tree where multiple nodes share the same horizontal distance and level. How should you modify the top view algorithm to always show the node with the smallest value at each horizontal distance?

hard

A. Store the last node encountered at each horizontal distance regardless of value

B. Use a stack instead of a queue for traversal

C. Ignore horizontal distance and print all nodes at the top level

D. Store the node with the smallest value for each horizontal distance, updating if a smaller value is found

Solution

1. Step 1: Understand the problem with duplicates

   When nodes share horizontal distance and level, top view must pick one node per horizontal distance.

2. Step 2: Modify map to track smallest value

   Instead of storing first node only, update stored node if a smaller value is found at the same horizontal distance.

3. Step 3: Implement comparison logic

   Check if current node's value is smaller than stored; if yes, replace it in the map.

4. Final Answer:

   Store the node with the smallest value for each horizontal distance, updating if a smaller value is found -> Option D

5. Quick Check:

   Keep smallest value per horizontal distance [OK]

Hint: Update map only if new node value is smaller [OK]

Common Mistakes:

• Ignoring node values when duplicates occur
• Using last node instead of smallest value
• Changing traversal order instead of map logic