Left Side View of Binary Tree in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Left Side View for a Simple Binary Tree

What is the output of the left side view function for the given binary tree?

DSA Python

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

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

# Tree construction
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))

A [1, 2, 5]

B [1, 2, 4]

C [1, 3, 5]

D [1, 3, 4]

💡 Hint

Remember, the left side view shows the first node visible at each level from the left side.

The left side view collects the first node at each level when traversing level by level from left to right. At level 0, node 1 is visible; at level 1, node 2 is the first; at level 2, node 4 is the first visible node.

🧠 Conceptual

intermediate

Understanding Left Side View Node Selection

In the left side view of a binary tree, which nodes are included in the output?

A All leaf nodes in the tree.

B All nodes that have a left child.

C The last node encountered at each level when traversing from right to left.

D The first node encountered at each level when traversing from left to right.

💡 Hint

Think about what you see when looking at the tree from the left side.

The left side view includes the first node visible at each level from the left side, which corresponds to the first node encountered at each level during a left-to-right level order traversal.

🔧 Debug

advanced

Identify the Error in Left Side View Implementation

What error will the following code produce when trying to get the left side view of a binary tree?

DSA Python

def left_side_view(root):
    result = []
    queue = [root]
    while queue:
        level_length = len(queue)
        for i in range(level_length):
            node = queue.pop()
            if i == 0:
                result.append(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
    return result

A TypeError because node is None

B IndexError due to popping from empty list

C Incorrect output because nodes are popped from the end instead of the front

D No error, correct output

💡 Hint

Check how nodes are removed from the queue and how it affects traversal order.

Using pop() without an index removes the last element, causing the traversal to process nodes in reverse order, which leads to incorrect left side view output.

🚀 Application

advanced

Left Side View of a Complete Binary Tree

Given a complete binary tree with nodes numbered 1 to 7 as shown below, what is the left side view output? 1 / \ 2 3 / \ / \ 4 5 6 7

A [1, 2, 4]

B [1, 3, 7]

C [1, 2, 5]

D [1, 3, 6]

💡 Hint

Look at the first node visible at each level from the left side.

At level 0, node 1 is visible; at level 1, node 2 is the first node; at level 2, node 4 is the first node visible from the left.

🧠 Conceptual

expert

Why Use Level Order Traversal for Left Side View?

Why is level order traversal (breadth-first) preferred to find the left side view of a binary tree?

A Because it processes nodes level by level, allowing easy selection of the first node at each level.

B Because it only visits left children of nodes.

C Because it visits nodes in pre-order sequence.

D Because it uses less memory than depth-first traversal.

💡 Hint

Think about how nodes are grouped by levels in the traversal.

Level order traversal visits nodes level by level, making it straightforward to pick the first node at each level for the left side view.

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