Why Left Side View of Binary Tree in DSA Java?

The Big Idea

Discover how to see a tree's shape from the left side without missing a single branch!

The Scenario

Imagine you have a tall tree in your backyard. You want to see which branches you can spot if you stand on the left side of the tree. But if you try to look at every branch from the top or from the front, it gets confusing and hard to tell which branches are visible from the left.

The Problem

Trying to find the left side view by checking every branch manually is slow and confusing. You might miss some branches or count the same branch multiple times because you don't have a clear way to look only from the left side. It's like trying to draw the tree from the left without a guide.

The Solution

The Left Side View of a Binary Tree helps you see exactly which nodes (branches) are visible when looking from the left side. It picks the first node you see at each level of the tree, making it easy to understand the tree's shape from that side without confusion.

Before vs After

✗ Before

void printLeftView(Node root) {
  // Manually check all nodes and guess which are visible
  // No clear method, leads to errors and missed nodes
}

✓ After

void printLeftView(Node root) {
  int[] maxLevel = {0};
  printLeftViewUtil(root, 1, maxLevel);
}

void printLeftViewUtil(Node node, int level, int[] maxLevel) {
  if (node == null) return;
  if (level > maxLevel[0]) {
    System.out.print(node.data + " ");
    maxLevel[0] = level;
  }
  printLeftViewUtil(node.left, level + 1, maxLevel);
  printLeftViewUtil(node.right, level + 1, maxLevel);
}

What It Enables

This concept lets you quickly see the outline of a tree from the left side, helping in visualizing and solving tree problems easily.

Real Life Example

In a family tree, the left side view can show the oldest ancestor at each generation level when viewed from the left, helping you understand the family lineage clearly.

Key Takeaways

Manual checking of visible nodes is confusing and error-prone.

Left Side View picks the first visible node at each level from the left.

This helps visualize the tree's shape clearly and simply.

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