Left Side View of Binary Tree in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Left Side View of a binary tree?

The left side view of a binary tree is the set of nodes visible when the tree is seen from the left side. It includes the leftmost node at each level of the tree.

beginner

Which tree traversal method is commonly used to find the left side view of a binary tree?

Level order traversal (Breadth-First Search) is commonly used because it processes nodes level by level, allowing us to pick the first node at each level as the leftmost node.

beginner

In the context of left side view, why do we pick the first node at each level during traversal?

Because the first node visited at each level from left to right is the leftmost node visible from the left side of the tree.

beginner

What data structure is typically used to implement level order traversal for left side view?

A queue is used to hold nodes of the current level while traversing the tree level by level.

beginner

How does the left side view differ from the right side view of a binary tree?

The left side view shows the leftmost nodes at each level, while the right side view shows the rightmost nodes at each level when viewed from the right side.

What is the first step to find the left side view of a binary tree?

A Perform preorder traversal

B Perform inorder traversal

C Perform postorder traversal

D Perform level order traversal

Level order traversal processes nodes level by level, which helps identify the leftmost node at each level.

Which data structure helps keep track of nodes at each level during traversal?

A Stack

B Array

C Queue

D HashMap

A queue is used in level order traversal to process nodes level by level.

In the left side view, which node at each level is included in the output?

A Leftmost node

B Last visited node

C Rightmost node

D Root node only

The leftmost node at each level is visible from the left side and included in the left side view.

If a level has nodes 4, 5, and 6 from left to right, which node appears in the left side view?

A 5

B 4

C 6

D None

Node 4 is the leftmost node at that level and will be visible from the left side.

What is the time complexity of finding the left side view using level order traversal?

A O(n)

B O(log n)

C O(n^2)

D O(1)

Each node is visited once, so the time complexity is O(n), where n is the number of nodes.

Explain how to find the left side view of a binary tree using level order traversal.

Think about how you see the tree from the left side and which nodes are visible.

Describe the difference between left side view and right side view of a binary tree.

Imagine standing on the left or right side of the tree.

Practice

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

easy

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

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

C. All leaf nodes of the tree

D. Nodes at the deepest level of the tree

Solution

1. Step 1: Understand the view perspective

   The left side view means looking at the tree from the left side, so only nodes visible from that side are included.

2. Step 2: Identify what nodes are visible

   Nodes that are the first in each level from the left are visible, so the leftmost nodes at each depth form the left side view.

3. Final Answer:

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

4. Quick Check:

   Left side view = nodes visible from left side [OK]

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

Common Mistakes:

• Confusing left side view with right side view
• Including all leaf nodes instead of leftmost nodes
• Thinking it shows deepest nodes only

2. Which Java data structure is best suited to help find the left side view of a binary tree level by level?

easy

A. Queue

B. Stack

C. HashMap

D. ArrayList

Solution

1. Step 1: Identify traversal method for levels

   Level order traversal visits nodes level by level, which is needed to find leftmost nodes at each level.

2. Step 2: Choose data structure for level order

   Queue is used for breadth-first search (level order traversal) because it processes nodes in FIFO order.

3. Final Answer:

   Queue -> Option A

4. Quick Check:

   Level order traversal uses queue [OK]

Hint: Use queue for level order traversal to get left side view [OK]

Common Mistakes:

• Using stack which is for depth-first traversal
• Using HashMap which stores key-value pairs, not order
• Using ArrayList without order control

3. What is the output of the left side view for this binary tree?

    1
   / \
  2   3
 /     \
4       5

medium

A. [1, 4, 5]

B. [1, 3, 5]

C. [1, 2, 5]

D. [1, 2, 4]

Solution

1. Step 1: Identify leftmost nodes at each level

   Level 0: Node 1 is leftmost.
   Level 1: Node 2 is leftmost.
   Level 2: Node 4 is leftmost.

2. Step 2: Collect nodes in order

   Left side view nodes are [1, 2, 4].

3. Final Answer:

   [1, 2, 4] -> Option D

4. Quick Check:

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

Hint: Pick first node at each level from left [OK]

Common Mistakes:

• Choosing right child nodes instead of left
• Including nodes not visible from left
• Mixing order of nodes

4. Given this Java code snippet for left side view, what is the error?

void leftSideView(Node root) {
  Queue<Node> q = new LinkedList<>();
  q.add(root);
  while (!q.isEmpty()) {
    int size = q.size();
    for (int i = 0; i < size; i++) {
      Node curr = q.poll();
      if (i == 0) System.out.print(curr.data + " ");
      if (curr.left != null) q.add(curr.left);
      if (curr.right != null) q.add(curr.right);
    }
  }
}

medium

A. Printing all nodes instead of only left side nodes

B. Using poll() instead of peek() to get current node

C. No check if root is null before adding to queue

D. Incorrect loop condition in for loop

Solution

1. Step 1: Check root handling

   The code adds root to queue without checking if root is null, which can cause NullPointerException.

2. Step 2: Verify other parts

   Using poll() is correct to remove and get node.
   Printing only when i==0 prints leftmost node.
   Loop condition is correct for level size.

3. Final Answer:

   No check if root is null before adding to queue -> Option C

4. Quick Check:

   Null root check missing causes error [OK]

Hint: Always check if root is null before processing [OK]

Common Mistakes:

• Ignoring null root causing runtime error
• Confusing poll() with peek() usage
• Printing all nodes instead of leftmost only

5. How would you modify the left side view algorithm to return the left side view nodes in reverse order (bottom to top)?

hard

A. Traverse the tree in reverse level order using a queue

B. Use a stack to store leftmost nodes during level order, then pop all

C. Use recursion to collect nodes and reverse the list at the end

D. Print nodes during traversal without storing them

Solution

1. Step 1: Understand normal left side view collection

   Normally, leftmost nodes are collected in order from top to bottom during level order traversal.

2. Step 2: Modify to reverse order

   Store leftmost nodes in a stack during traversal, then pop all to get bottom-to-top order.

3. Final Answer:

   Use a stack to store leftmost nodes during level order, then pop all -> Option B

4. Quick Check:

   Stack reverses order of collected nodes [OK]

Hint: Use stack to reverse order of collected left side nodes [OK]

Common Mistakes:

• Trying to reverse queue order directly
• Not storing nodes before printing
• Using recursion without reversing result