Why Left Side View of Binary Tree in DSA Python?

The Big Idea

What if you could instantly see the 'front row' of a complex tree without missing a single branch?

The Scenario

Imagine you have a tall tree in your backyard, and you want to see which branches are visible if you stand on the left side. If you try to list these branches by looking at the tree yourself, it can be confusing and easy to miss some branches hidden behind others.

The Problem

Manually checking each branch from the left side is slow and error-prone because you might forget to look at some branches or get confused about which ones are actually visible. It's hard to keep track of the order and depth of branches without a clear method.

The Solution

The Left Side View of a Binary Tree helps by giving a clear way to find and list only the branches visible from the left side. It uses a simple step-by-step process to check each level of the tree and pick the first branch seen from the left, making it easy and reliable.

Before vs After

✗ Before

def left_view_manual(tree):
    # Try to guess visible nodes by random checks
    visible = []
    # No clear method, might miss nodes
    return visible

✓ After

def left_view(tree):
    result = []
    def helper(node, level):
        if not node:
            return
        if level == len(result):
            result.append(node.val)
        helper(node.left, level + 1)
        helper(node.right, level + 1)
    helper(tree, 0)
    return result

What It Enables

This concept makes it easy to see the important parts of a tree from one side, enabling better understanding and visualization of tree structures.

Real Life Example

In a family tree, the left side view can show the oldest ancestors visible from one side, helping you quickly see the main family line without confusion.

Key Takeaways

Manual checking of visible nodes is confusing and unreliable.

Left Side View uses a clear method to find visible nodes level by level.

This helps visualize and understand tree structures easily.

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