Left Side View of Binary Tree in DSA Java - Build from Scratch

Left Side View of Binary Tree

📖 Scenario: You are working with a binary tree data structure that represents a family tree. You want to see the names of the family members visible from the left side of the tree.

🎯 Goal: Build a Java program that creates a binary tree, sets a maximum depth to explore, finds the left side view of the tree, and prints the visible nodes from left to right.

📋 What You'll Learn

Create a binary tree with specific nodes and values

Add a variable to track the maximum depth visited

Write a recursive function to find the left side view nodes

Print the left side view nodes in order

💡 Why This Matters

🌍 Real World

Left side view of a binary tree helps visualize hierarchical data from a specific perspective, useful in family trees, organizational charts, and UI layouts.

💼 Career

Understanding tree traversal and views is important for software engineers working with data structures, algorithms, and system design.

1

Create the Binary Tree

Create a class called TreeNode with int val, TreeNode left, and TreeNode right fields. Then create a binary tree with root node value 1, left child 2, right child 3, left child of 2 as 4, and right child of 2 as 5.

DSA Java

// Define TreeNode class and create the binary tree
// Your code here

Hint

Define the TreeNode class first, then create the nodes and link them as described.

2

Add Maximum Depth Tracker

Inside the Main class, add a static integer variable called maxDepth and initialize it to 0. This will track the maximum depth visited during traversal.

DSA Java

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

public class Main {
    static int maxDepth = 0;

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);
        // Your code here
    }
}

Hint

Add the maxDepth variable inside the Main class but outside the main method.

3

Write Recursive Function for Left Side View

Add a static method called leftSideView that takes TreeNode node, int depth, and a List<Integer> result. If node is null, return. If depth is greater than maxDepth, add node.val to result and update maxDepth. Then recursively call leftSideView on node.left and node.right with depth + 1.

DSA Java

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

import java.util.*;

public class Main {
    static int maxDepth = 0;

    public static void leftSideView(TreeNode node, int depth, List result) {
        // Your code here
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);
    }
}

Hint

Use a recursive method that adds the first node at each depth to the result list.

4

Print the Left Side View

In the main method, create a List<Integer> called leftView. Call leftSideView(root, 1, leftView). Then print the leftView list.

DSA Java

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

import java.util.*;

public class Main {
    static int maxDepth = 0;

    public static void leftSideView(TreeNode node, int depth, List result) {
        if (node == null) {
            return;
        }
        if (depth > maxDepth) {
            result.add(node.val);
            maxDepth = depth;
        }
        leftSideView(node.left, depth + 1, result);
        leftSideView(node.right, depth + 1, result);
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);

        List leftView = new ArrayList();
        // Your code here
    }
}

Hint

Call the recursive function starting at depth 1 and print the list after the call.

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