Why Top View of Binary Tree in DSA Python?

The Big Idea

What if you could instantly see which parts of a complex tree stand out from above without any guesswork?

The Scenario

Imagine you have a tall tree in your backyard and you want to see which branches are visible if you look at it from the top. If you try to list these branches by just guessing or drawing without a clear method, it becomes confusing and you might miss some branches or count some twice.

The Problem

Trying to find the top view of a tree manually is slow and error-prone because you have to consider which branches overlap when seen from above. Without a systematic way, you might miss branches hidden behind others or get lost in the tree's complexity.

The Solution

The top view of a binary tree uses a smart way to track the horizontal positions of nodes and picks only the first node visible at each position. This method quickly and clearly shows which nodes are visible from the top, avoiding confusion and mistakes.

Before vs After

✗ Before

def print_top_view(root):
    # Manually guess visible nodes
    print(root.data)
    # No clear way to handle overlaps or positions

✓ After

def print_top_view(root):
    # Use horizontal distance and level tracking
    # Store first node at each horizontal distance
    # Print nodes in order of horizontal distance

What It Enables

This concept lets you clearly see the outline of a tree from above, which helps in understanding tree structure and solving problems related to visibility and layout.

Real Life Example

In city planning, when designing roads and buildings, planners need to know which buildings are visible from above to avoid blocking views or sunlight. The top view concept helps model such visibility clearly.

Key Takeaways

Manual guessing of top view is confusing and error-prone.

Top view uses horizontal positions to pick visible nodes.

This method gives a clear, fast way to see the tree's outline from above.

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