Top View of Binary Tree in DSA Python - 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.

Specifically, how does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

from collections import deque

def top_view(root):
    if not root:
        return []
    queue = deque([(root, 0)])
    top_nodes = {}
    while queue:
        node, hd = queue.popleft()
        if hd not in top_nodes:
            top_nodes[hd] = node.val
        if node.left:
            queue.append((node.left, hd - 1))
        if node.right:
            queue.append((node.right, hd + 1))
    return [top_nodes[hd] for hd in sorted(top_nodes)]
    

This code finds the top view of a binary tree by visiting nodes level by level and tracking their horizontal distances.

Identify Repeating Operations

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

How Execution Grows With Input

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

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 in a straight line as the tree grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the top view grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The code visits nodes multiple times, so it must be slower than O(n)."

[OK] Correct: Each node is visited exactly once in the queue, so the total work is proportional to the number of nodes.

Interview Connect

Understanding this helps you explain how tree traversals work efficiently and how to track extra information like horizontal distance.

Self-Check

"What if we used a recursive approach instead of a queue? How would the time complexity change?"

Practice

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

easy

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

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

C. All leaf nodes of the tree

D. Nodes visible when the tree is viewed from above

Solution

1. Step 1: Understand the meaning of top view

   The top view of a binary tree shows nodes that are visible when looking down vertically from above the tree.

2. Step 2: Compare with other views

   Left or right views show nodes visible from sides, and leaf nodes are only the bottom nodes, so they are different.

3. Final Answer:

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

4. Quick Check:

   Top view = nodes seen from above [OK]

Hint: Top view means looking down from above the tree [OK]

Common Mistakes:

• Confusing top view with left or right view
• Thinking top view shows all leaf nodes
• Assuming top view includes all nodes

2. Which of the following Python data structures is best suited to store nodes by their horizontal distance for top view calculation?

easy

A. Set

B. List

C. Dictionary

D. Tuple

Solution

1. Step 1: Identify the need for key-value mapping

   We need to map horizontal distances (keys) to nodes (values) to track the first node at each horizontal distance.

2. Step 2: Choose the appropriate data structure

   Dictionaries provide key-value pairs, allowing efficient lookup and insertion by horizontal distance.

3. Final Answer:

   Dictionary -> Option C

4. Quick Check:

   Use dictionary for horizontal distance mapping [OK]

Hint: Use dictionary to map horizontal distance to nodes [OK]

Common Mistakes:

• Using list which is index-based, not key-based
• Using set which stores unique values without keys
• Using tuple which is immutable and not key-value

3. Consider the following binary tree:

        1
       / \
      2   3
       \   \
        4   5

What is the top view of this tree?

medium

A. [2, 1, 3, 5]

B. [1, 2, 4, 5]

C. [2, 1, 3]

D. [1, 4, 5]

Solution

1. Step 1: Assign horizontal distances

   Root (1) is at 0, left child (2) at -1, right child (3) at +1, node 4 at 0 (right of 2), node 5 at +2 (right of 3).

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

   At -1: 2, at 0: 1 (4 is below 1), at +1: 3, at +2: 5.

3. Final Answer:

   [2, 1, 3, 5] -> Option A

4. Quick Check:

   Top view nodes by horizontal distance = [2,1,3,5] [OK]

Hint: Pick first node at each horizontal distance from left to right [OK]

Common Mistakes:

• Including node 4 instead of node 1 at horizontal distance 0
• Missing node 5 at horizontal distance +2
• Confusing order of nodes in output

4. The following code snippet is intended to print the top view of a binary tree but has a bug. What is the error?

from collections import deque

def top_view(root):
    if not root:
        return []
    q = deque([(root, 0)])
    hd_node = {}
    while q:
        node, hd = q.pop()
        if hd not in hd_node:
            hd_node[hd] = node.data
        if node.left:
            q.append((node.left, hd - 1))
        if node.right:
            q.append((node.right, hd + 1))
    return [hd_node[hd] for hd in sorted(hd_node)]

medium

A. Using q.pop() instead of q.popleft() causes wrong traversal order

B. hd_node dictionary is not initialized correctly

C. The horizontal distance is not updated properly

D. The function does not handle empty trees

Solution

1. Step 1: Analyze queue operations

   The code uses deque as a queue but uses q.pop(), which removes from the right (stack behavior), not from the left (queue behavior).

2. Step 2: Understand traversal order impact

   Top view requires level order traversal (BFS), so nodes must be processed in FIFO order using q.popleft(). Using pop() reverses order and causes incorrect results.

3. Final Answer:

   Using q.pop() instead of q.popleft() causes wrong traversal order -> Option A

4. Quick Check:

   Use popleft() for BFS queue traversal [OK]

Hint: Use popleft() for BFS, pop() is for stack (LIFO) [OK]

Common Mistakes:

• Confusing pop() with popleft() in deque
• Not realizing BFS needs FIFO order
• Thinking dictionary initialization is the problem

5. Given a binary tree where multiple nodes share the same horizontal distance and level, which node should be included in the top view?

hard

A. The node with the maximum value at that horizontal distance

B. The node that appears first in level order traversal at that horizontal distance

C. The node that appears last in level order traversal at that horizontal distance

D. All nodes at that horizontal distance

Solution

1. Step 1: Understand top view node selection

   Top view includes the first node encountered at each horizontal distance during level order traversal (BFS), which means the node closest to the root vertically.

2. Step 2: Eliminate other options

   Maximum value or last node in traversal is not considered; all nodes at same horizontal distance are not included, only the first one.

3. Final Answer:

   The node that appears first in level order traversal at that horizontal distance -> Option B

4. Quick Check:

   Top view = first node at horizontal distance in BFS [OK]

Hint: Pick first node at horizontal distance in BFS order [OK]

Common Mistakes:

• Choosing node by value instead of position
• Including all nodes at same horizontal distance
• Picking last node instead of first in BFS