Bottom View of Binary Tree in DSA Python - Time & Space Complexity

Time Complexity: Bottom View of Binary Tree

O(n)

Understanding Time Complexity

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

How does the work increase when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

from collections import deque

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

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• 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 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
100	About 100 visits
1000	About 1000 visits

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 bottom view grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The time complexity is more than linear because we sort the horizontal distances at the end."

[OK] Correct: Sorting happens on the horizontal distances, which are at most the width of the tree, usually much smaller than n, so it does not change the overall linear time complexity.

Interview Connect

Understanding how to analyze tree traversals like this helps you explain your approach clearly and shows you can reason about efficiency in real problems.

Self-Check

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

Practice

1. What does the bottom 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 top

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

D. Nodes visible when the tree is viewed from the bottom

Solution

1. Step 1: Understand the bottom view concept

   The bottom view shows nodes visible when looking at the tree from below, meaning the lowest nodes at each horizontal distance.

2. Step 2: Compare with other views

   Top view shows nodes from above, left and right views show nodes from sides, so bottom view is unique for bottom perspective.

3. Final Answer:

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

4. Quick Check:

   Bottom view = visible nodes from below [OK]

Hint: Bottom view shows lowest nodes at each horizontal distance [OK]

Common Mistakes:

• Confusing bottom view with top view
• Thinking bottom view shows all leaf nodes
• Mixing bottom view with side views

2. Which data structure is commonly used to track nodes and their horizontal distances when computing the bottom view of a binary tree?

easy

A. Hash Map (Dictionary)

B. Queue

C. Stack

D. Linked List

Solution

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

   We need to map horizontal distances to nodes to know which node is visible at each position.

2. Step 2: Choose the data structure

   A hash map (dictionary) stores horizontal distance as key and node value as value, updating with the bottom-most node.

3. Final Answer:

   Hash Map (Dictionary) -> Option A

4. Quick Check:

   Use dictionary to map horizontal distance to node [OK]

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

Common Mistakes:

• Using stack which is LIFO and not suitable here
• Using queue alone without mapping distances
• Using linked list which is inefficient for key-value mapping

3. What is the bottom view output of the following binary tree?

        20
       /  \
     8     22
    / \      \
   5   3      25
      / \     
     10  14

medium

A. [5, 10, 3, 14, 25]

B. [5, 8, 10, 14, 25]

C. [5, 10, 3, 22, 25]

D. [8, 10, 3, 14, 22]

Solution

1. Step 1: Assign horizontal distances and find bottom nodes

   Horizontal distances: 20(0), 8(-1), 22(1), 5(-2), 3(0), 25(2), 10(-1), 14(1). Bottom nodes at each distance: -2:5, -1:10, 0:3, 1:14, 2:25.

2. Step 2: List bottom view nodes in order of horizontal distance

   From left to right: 5, 10, 3, 14, 25.

3. Final Answer:

   [5, 10, 3, 14, 25] -> Option A

4. Quick Check:

   Bottom view nodes by horizontal distance = [5,10,3,14,25] [OK]

Hint: Track horizontal distances and update bottom nodes [OK]

Common Mistakes:

• Not updating bottom node when a lower node appears
• Mixing horizontal distances order
• Including nodes not visible from bottom

4. Consider this code snippet to compute bottom view of a binary tree:

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

What is the main error in this code?

medium

A. Not checking if root is None

B. Using pop() removes last element, causing wrong traversal order

C. Not updating hd_node dictionary

D. Incorrect horizontal distance calculation

Solution

1. Step 1: Analyze queue usage and traversal order

   The code uses queue.pop() which removes last element (LIFO), making it behave like a stack, causing depth-first traversal instead of breadth-first.

2. Step 2: Understand impact on bottom view calculation

   Breadth-first traversal is needed to process nodes level by level to correctly update bottom nodes. Using pop() breaks this order.

3. Final Answer:

   Using pop() removes last element, causing wrong traversal order -> Option B

4. Quick Check:

   Use pop(0) or deque.popleft() for BFS [OK]

Hint: Use pop(0) for queue to maintain BFS order [OK]

Common Mistakes:

• Using pop() instead of pop(0) for queue
• Ignoring empty root check
• Incorrect horizontal distance updates

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

hard

A. Store nodes in a list for each horizontal distance and pick the smallest value

B. Store only the first node encountered at each horizontal distance

C. Store both node value and depth in the dictionary, update only if new node depth is greater

D. Ignore depth and update dictionary with last visited node

Solution

1. Step 1: Understand the need to track depth

   To show the bottom-most node at each horizontal distance, we must know which node is deepest (greatest depth).

2. Step 2: Modify dictionary to store value and depth

   Store a tuple (node value, depth) for each horizontal distance. Update only if new node's depth is greater than stored depth.

3. Final Answer:

   Store both node value and depth in the dictionary, update only if new node depth is greater -> Option C

4. Quick Check:

   Track depth to update bottom view nodes correctly [OK]

Hint: Track depth with node to update bottom view correctly [OK]

Common Mistakes:

• Ignoring depth and overwriting nodes incorrectly
• Storing all nodes without filtering by depth
• Picking first or last node without depth check