Top View of Binary Tree in DSA Go - Time & Space Complexity

Time Complexity: Top View of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the top view of a binary tree changes as the tree grows.

How does the number of nodes affect the work done to get the top view?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

    result := []int{}
    for i := minHd; i <= maxHd; i++ {
        result = append(result, hdMap[i])
    }
    return result
}
    

This code finds the top view of a binary tree by doing a level order traversal and tracking horizontal distances.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that processes each node once in a queue.
• How many times: Exactly once per node in the tree (n times).

How Execution Grows With Input

As the number of nodes grows, the code visits each node once, so the work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits and checks
100	About 100 visits and checks
1000	About 1000 visits and checks

Pattern observation: The operations increase linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the top view grows in direct proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "The code visits nodes multiple times, so the time is more than linear."

[OK] Correct: Each node is added to the queue once and processed once, so the total work is proportional to the number of nodes.

Interview Connect

Understanding this linear time complexity helps you explain how tree traversals work efficiently and shows you can analyze common tree problems clearly.

Self-Check

"What if we used a depth-first traversal instead of breadth-first? How would the time complexity change?"

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