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

Time Complexity: Left Side View of Binary Tree

O(n)

Understanding Time Complexity

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

Specifically, how does the number of steps grow when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

from collections import deque

def left_side_view(root):
    if not root:
        return []
    result = []
    queue = deque([root])
    while queue:
        level_length = len(queue)
        for i in range(level_length):
            node = queue.popleft()
            if i == 0:
                result.append(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
    return result
    

This code finds the left side view of a binary tree by visiting nodes level by level and recording the first node at each level.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once in a breadth-first manner using a queue.
• How many times: Each node is processed exactly one time.

How Execution Grows With Input

As the number of nodes in the tree increases, the number of steps grows roughly the same way.

Input Size (n)	Approx. Operations
10	About 10 visits to nodes
100	About 100 visits to nodes
1000	About 1000 visits to nodes

Pattern observation: The work grows linearly with the number of nodes because each node is visited once.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the left side view grows directly in proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "The time complexity is O(log n) because we only look at one node per level."

[OK] Correct: Even though we record one node per level, we still visit all nodes to find that first node. So the total work depends on all nodes, not just one per level.

Interview Connect

Understanding this helps you explain how tree traversals work and how to analyze their efficiency, a common skill interviewers look for.

Self-Check

"What if we changed the traversal to depth-first search instead of breadth-first? How would the time complexity change?"

Practice

1. What does the Left Side View of a binary tree represent?

easy

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

B. Nodes at the deepest level of the tree

C. All leaf nodes of the tree

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

Solution

1. Step 1: Understand the view perspective

   The left side view shows nodes visible when looking at the tree from the left side.

2. Step 2: Identify visible nodes

   Only the first node at each level from the left is visible in this view.

3. Final Answer:

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

4. Quick Check:

   Left side view = nodes visible from left [OK]

Hint: Left side view shows first node at each level from left [OK]

Common Mistakes:

• Confusing left side view with right side view
• Thinking it shows all leaf nodes
• Assuming it shows deepest nodes only

2. Which of the following Python code snippets correctly initializes an empty list to store the left side view nodes?

easy

A. left_view = ()

B. left_view = {}

C. left_view = []

D. left_view = set()

Solution

1. Step 1: Identify the correct data structure

   We need a list to store nodes in order, so an empty list is best.

2. Step 2: Check syntax for list initialization

   Using square brackets [] creates an empty list in Python.

3. Final Answer:

   left_view = [] -> Option C

4. Quick Check:

   Empty list = [] [OK]

Hint: Use [] to create empty list for ordered nodes [OK]

Common Mistakes:

• Using {} which creates a dictionary, not a list
• Using () which creates a tuple, not a list
• Using set() which is unordered

3. What is the output of the following code that prints the left side view of the given binary tree?

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def left_side_view(root):
    result = []
    def dfs(node, level):
        if not node:
            return
        if level == len(result):
            result.append(node.val)
        dfs(node.left, level + 1)
        dfs(node.right, level + 1)
    dfs(root, 0)
    return result

root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.right = Node(4)
root.right.right = Node(5)

print(left_side_view(root))

medium

A. [1, 2, 3]

B. [1, 2, 4]

C. [1, 3, 5]

D. [1, 4, 5]

Solution

1. Step 1: Trace DFS traversal and level checks

   At level 0, node 1 is added. At level 1, node 2 is added (left child). At level 2, node 4 is added (left child's right child).

2. Step 2: Confirm nodes added only once per level

   Right children at each level are ignored if left child exists at that level.

3. Final Answer:

   [1, 2, 4] -> Option B

4. Quick Check:

   Left side view nodes = [1, 2, 4] [OK]

Hint: Add node if level equals result length during DFS [OK]

Common Mistakes:

• Adding right child nodes before left child nodes
• Appending nodes multiple times per level
• Confusing left and right children order

4. Find the error in the following code that attempts to print the left side view of a binary tree:

def left_side_view(root):
    result = []
    def dfs(node, level):
        if node is None:
            return
        if level < len(result):
            result.append(node.val)
        dfs(node.left, level + 1)
        dfs(node.right, level + 1)
    dfs(root, 0)
    return result

medium

A. The condition should be 'level == len(result)', not 'level < len(result)'

B. The base case should check 'if node' instead of 'if node is None'

C. The recursive calls should swap left and right children

D. The function should return 'result' inside the dfs function

Solution

1. Step 1: Analyze the condition for adding nodes

   The code uses 'level < len(result)' which is incorrect because nodes should be added only when level equals the current result length.

2. Step 2: Correct the condition to add nodes once per level

   Changing condition to 'level == len(result)' ensures only the first node at each level is added.

3. Final Answer:

   The condition should be 'level == len(result)', not 'level < len(result)' -> Option A

4. Quick Check:

   Use '==' to add first node per level [OK]

Hint: Add node only when level equals result length [OK]

Common Mistakes:

• Using '<' instead of '==' in condition
• Returning result inside recursive function
• Swapping left and right child calls unnecessarily

5. Given a binary tree, how would you modify the standard left side view algorithm to return the left side view nodes in reverse order (bottom to top)?

hard

A. Perform the usual left side view traversal, then reverse the result list before returning

B. Traverse the tree from right to left instead of left to right

C. Add nodes to the result list only if they are leaf nodes

D. Use a queue and perform level order traversal from right to left

Solution

1. Step 1: Perform standard left side view traversal

   Use DFS or BFS to collect the first node at each level from left to right.

2. Step 2: Reverse the collected list

   After traversal, reverse the list to get nodes from bottom to top.

3. Final Answer:

   Perform the usual left side view traversal, then reverse the result list before returning -> Option A

4. Quick Check:

   Reverse after left side view traversal [OK]

Hint: Reverse the left side view list after traversal [OK]

Common Mistakes:

• Changing traversal order instead of reversing result
• Collecting leaf nodes only
• Using right to left traversal which gives wrong nodes