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

Time Complexity: Left Side View of Binary Tree

O(n)

Understanding Time Complexity

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

How does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public List leftSideView(TreeNode root) {
    List result = new ArrayList<>();
    if (root == null) return result;
    Queue queue = new LinkedList<>();
    queue.offer(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        for (int i = 0; i < size; i++) {
            TreeNode node = queue.poll();
            if (i == 0) result.add(node.val);
            if (node.left != null) queue.offer(node.left);
            if (node.right != null) queue.offer(node.right);
        }
    }
    return result;
}
    

This code finds the left side view by visiting nodes level by level and picking 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.
• How many times: Each node is processed exactly one time.

How Execution Grows With Input

As the number of nodes increases, the work grows directly with the number of nodes because each node is visited once.

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

Pattern observation: The operations grow linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Since we only pick one node per level, the time is O(height) or O(log n)."

[OK] Correct: Even though we pick one node per level for the result, we still visit all nodes to know which is first at each level. So the total work depends on all nodes, not just the height.

Interview Connect

Understanding this helps you explain how tree traversal costs grow and shows you can analyze level-by-level processing clearly.

Self-Check

"What if we used a depth-first search instead of breadth-first search? 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 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