Left Side View of Binary Tree in DSA Rust - Deep Dive

Overview - Left Side View of Binary Tree

What is it?

The left side view of a binary tree is the set of nodes visible when you look at the tree from its left side. It means you see the first node at each level starting from the root down to the leaves. This view helps us understand the structure of the tree from a different angle. It is a way to extract specific nodes that represent the leftmost path at every depth.

Why it matters

Without the left side view, we might miss important information about the tree's shape and how nodes are arranged on the left. This view helps in visualizing and debugging tree structures, especially in problems where the leftmost nodes have special significance. It also aids in algorithms that need to process or display trees level by level from a particular perspective.

Where it fits

Before learning this, you should understand what a binary tree is and how to traverse it using methods like breadth-first search (BFS) or depth-first search (DFS). After this, you can explore other tree views like the right side view, top view, or bottom view, and advanced tree algorithms.

Mental Model

Core Idea

The left side view of a binary tree is the first node you encounter at each level when looking from the left side.

Think of it like...

Imagine standing on the left side of a tall bookshelf with many rows. The left side view is like seeing only the first book on each shelf from your side, ignoring the books behind it.

Binary Tree Levels and Left Side View:

Level 0:        1
               /   \
Level 1:     2       3
             / \     / \
Level 2:   4   5   6   7

Left Side View: 1 -> 2 -> 4 -> null

Build-Up - 6 Steps

1

Foundation

Understanding Binary Tree Structure

Concept: Learn what a binary tree is and how nodes connect.

A binary tree is a structure where each node has up to two children: left and right. The top node is called the root. Each child can have its own children, forming levels. For example, the root is level 0, its children are level 1, and so on. ```rust #[derive(Debug)] pub struct TreeNode { pub val: i32, pub left: Option>, pub right: Option>, } ```

Result

You can visualize a tree as layers of nodes connected by branches.

Understanding the tree's shape is essential before extracting any specific view or traversal.

2

Foundation

Level Order Traversal Basics

Concept: Learn how to visit nodes level by level using a queue.

Level order traversal visits nodes from top to bottom, left to right at each level. We use a queue to keep track of nodes to visit next. Start with the root, then enqueue its children, and repeat until all nodes are visited. ```rust use std::collections::VecDeque; let mut queue = VecDeque::new(); if let Some(root) = root { queue.push_back(root); } while !queue.is_empty() { let node = queue.pop_front().unwrap(); // process node if let Some(ref left) = node.left { queue.push_back(left.clone()); } if let Some(ref right) = node.right { queue.push_back(right.clone()); } } ```

Result

Nodes are visited in order: 1, 2, 3, 4, 5, 6, 7 for the example tree.

Level order traversal helps us process nodes level-wise, which is key to finding the left side view.

3

Intermediate

Identifying Leftmost Nodes at Each Level

🤔Before reading on: do you think the left side view includes all nodes on the left branch or only the first node at each level? Commit to your answer.

Concept: The left side view includes only the first node encountered at each level during level order traversal.

When doing level order traversal, the first node you dequeue at each level is the leftmost node visible from the left side. By recording these nodes, you get the left side view.

Result

Left side view nodes for the example: 1, 2, 4.

Knowing that only the first node per level matters simplifies the problem and avoids unnecessary checks.

4

Intermediate

Implementing Left Side View Using BFS in Rust

🤔Before reading on: do you think using a queue is better than recursion for left side view? Commit to your answer.

Concept: Use a queue to perform BFS and capture the first node at each level in Rust.

Create a queue and push the root node. Process each level by tracking its size. Record the value of the first node (index 0) in each level. Enqueue left then right children. ```rust use std::collections::VecDeque; pub fn left_side_view(root: Option>) -> Vec { let mut result = Vec::new(); let mut queue: VecDeque> = VecDeque::new(); if let Some(node) = root { queue.push_back(node); } while !queue.is_empty() { let level_size = queue.len(); for i in 0..level_size { let node = queue.pop_front().unwrap(); if i == 0 { result.push(node.val); } if let Some(left) = node.left { queue.push_back(left); } if let Some(right) = node.right { queue.push_back(right); } } } result } ```

Result

Output vector contains [1, 2, 4] for the example tree.

Using BFS with a queue matches the natural level order and makes capturing leftmost nodes straightforward.

5

Advanced

Recursive DFS Approach for Left Side View

🤔Before reading on: do you think DFS can find the left side view as easily as BFS? Commit to your answer.

Concept: Use depth-first search with tracking of levels to record the first node visited at each depth.

Use a helper function that takes current node, level, and mutable result vector. Record node if level equals current result length (first visit). Recurse left first, then right. ```rust pub fn left_side_view_dfs(root: Option>) -> Vec { let mut result = Vec::new(); dfs(&root, 0, &mut result); result } fn dfs(node: &Option>, level: usize, result: &mut Vec) { if let Some(n) = node { if level == result.len() { result.push(n.val); } dfs(&n.left, level + 1, result); dfs(&n.right, level + 1, result); } } ```

Result

Left side view nodes collected recursively: [1, 2, 4].

DFS can also find the left side view by careful ordering and level tracking, offering an alternative to BFS.

6

Expert

Handling Edge Cases and Large Trees Efficiently

🤔Before reading on: do you think the left side view changes if the tree is skewed or unbalanced? Commit to your answer.

Concept: Consider trees with missing children, skewed shapes, or very large sizes and optimize traversal accordingly.

Handle empty tree (return empty vec). For skewed left: all nodes included. Skewed right: root + right chain if no lefts. For large trees, BFS queue peaks at max width; DFS recursion depth is height (prefer for deep trees). Always check Option. ```rust // Edge case: empty tree assert_eq!(left_side_view(None), vec![]); // Skewed left // 1 // / // 2 // / //3 -> [1,2,3] ```

Result

Correct left side view regardless of tree shape, with efficient memory and time use.

Understanding tree shape and traversal limits helps write robust, efficient code for real-world trees.

Under the Hood

The left side view is found by traversing the tree level by level and selecting the first node encountered at each level. Internally, a queue stores nodes to visit next. For BFS, nodes are dequeued in order, and their children enqueued. For DFS, recursion tracks the current depth and records the first node visited at that depth. Memory holds the nodes seen so far to avoid duplicates.

Why designed this way?

This approach leverages the natural layering of trees. BFS naturally processes nodes level-wise, making it easy to pick the first node per level. DFS requires careful ordering but can be more memory efficient in some cases. Alternatives like storing all nodes and filtering later are less efficient. The design balances simplicity, clarity, and performance.

Queue-based BFS for Left Side View:

[Start]
  ↓
[Enqueue Root]
  ↓
[While queue not empty]
  ├─ Dequeue node (first at level) -> Record value
  ├─ Enqueue left child if exists
  └─ Enqueue right child if exists
  ↓
[Repeat for next level]
  ↓
[Result: Left side view nodes]

Myth Busters - 3 Common Misconceptions

Quick: Does the left side view include all nodes on the left branch? Commit yes or no.

Common Belief: The left side view includes every node on the left branch of the tree.

Reality: The left side view includes only the first node visible at each level, not all nodes on the left branch.

Why it matters: Mistaking this leads to incorrect outputs and confusion about what the left side view represents.

Quick: Can the right child ever appear in the left side view? Commit yes or no.

Common Belief: Only left children appear in the left side view; right children are never included.

Reality: Right children can appear in the left side view if the left child is missing at that level.

Why it matters: Ignoring right children causes missing nodes in the view, leading to incomplete or wrong results.

Quick: Is DFS always better than BFS for left side view? Commit yes or no.

Common Belief: DFS is always better than BFS for finding the left side view because it uses less memory.

Reality: Both DFS and BFS have tradeoffs; BFS is often simpler and clearer for level-based views, while DFS requires careful level tracking.

Why it matters: Choosing the wrong method without understanding tradeoffs can cause inefficient or complex code.

Expert Zone

1

The order of child traversal in DFS (left before right) is critical to correctly capture the left side view.

2

In BFS, the queue size can grow large for wide trees, so memory management matters in production.

3

Handling null or missing children carefully prevents bugs in both BFS and DFS implementations.

When NOT to use

Left side view is not suitable when you need full tree traversal or views from other angles like right, top, or bottom. For those, use corresponding algorithms like right side view or vertical order traversal.

Production Patterns

In real systems, left side view helps in UI tree visualizations, debugging tree structures, and in algorithms that prioritize leftmost nodes, such as certain search optimizations or hierarchical data displays.

Connections

Right Side View of Binary Tree

Opposite perspective of the same problem, focusing on rightmost nodes instead of leftmost.

Understanding left side view helps grasp right side view by symmetry, reinforcing level order traversal concepts.

Breadth-First Search (BFS)

Left side view uses BFS to process nodes level by level.

Mastering BFS is essential to efficiently extract views that depend on tree levels.

Human Visual Perception

Both involve selecting visible elements from a viewpoint, filtering out hidden parts.

Knowing how humans perceive layers and occlusion helps understand why only certain nodes appear in the left side view.

Common Pitfalls

#1 Recording all nodes at each level instead of only the first.

Wrong approach:

```rust
// Pseudo - would need full level collection first
let mut levels = vec![];
// ... collect all levels
for level in levels {
for node in level {
result.push(node.val);
}
}
```

Correct approach:

```rust
while !queue.is_empty() {
let level_size = queue.len();
for i in 0..level_size {
let node = queue.pop_front().unwrap();
if i == 0 {
result.push(node.val);
}
// enqueue children
}
}
```

Root cause:

Misunderstanding that the left side view requires only the first node per level, not all nodes.

#2 Ignoring right children when left child is missing.

Wrong approach:

```rust
if let Some(left) = node.left {
queue.push_back(left);
} // forgets right
```

Correct approach:

```rust
if let Some(left) = node.left {
queue.push_back(left);
}
if let Some(right) = node.right {
queue.push_back(right);
}
```

Root cause:

Assuming left child always exists and forgetting right child can be visible from left side.

#3 Using DFS without tracking levels properly.

Wrong approach:

```rust
fn dfs(node: &Option>) {
if node.is_none() { return; }
let n = node.as_ref().unwrap();
result.push(n.val); // pushes every node
dfs(&n.left);
dfs(&n.right);
}
```

Correct approach:

```rust
fn dfs(node: &Option>, level: usize, result: &mut Vec) {
if let Some(n) = node {
if level == result.len() {
result.push(n.val);
}
dfs(&n.left, level + 1, result);
dfs(&n.right, level + 1, result);
}
}
```

Root cause:

Not tracking levels causes all nodes to be added, losing the left side view logic.

Key Takeaways

The left side view shows the first visible node at each level when looking from the left side.

Level order traversal (BFS) naturally fits this problem by processing nodes level by level.

Only the first node at each level is included, not all nodes on the left branch.

DFS can also find the left side view by visiting left children first and tracking levels.

Handling missing children and tree shapes carefully ensures correct and efficient results.

Practice

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

easy

A. All leaf nodes of the tree

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

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

D. Nodes at the deepest level of the tree

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 C

4. Quick Check:

   Left side view = nodes seen from left [OK]

Hint: Think of standing on the left side and looking at the tree [OK]

Common Mistakes:

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

2. Which Rust code snippet correctly declares a binary tree node struct for left side view traversal?

easy

A. struct TreeNode { val: i32, left: TreeNode, right: TreeNode }

B. struct TreeNode { val: i32, left: Box<TreeNode>, right: Box<TreeNode> }

C. struct TreeNode { val: i32, left: &TreeNode, right: &TreeNode }

D. struct TreeNode { val: i32, left: Option, right: Option }

Solution

1. Step 1: Check ownership and optional children

   Rust binary trees use Option for nullable child nodes.

2. Step 2: Validate correct syntax

   Option allows for null or Some(child), which is correct for left and right.

3. Final Answer:

   struct TreeNode { val: i32, left: Option, right: Option } -> Option D

4. Quick Check:

   Use Option for nullable children [OK]

Hint: Use Option for child nodes in Rust [OK]

Common Mistakes:

• Using direct TreeNode type for children (causes infinite size)
• Using references without lifetimes
• Omitting Option for nullable children

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

    1
   / \
  2   3
   \   \
    5   4

Assuming the function returns a vector of node values visible from the left side.

medium

A. [1, 3, 4]

B. [1, 2, 5]

C. [1, 2, 4]

D. [1, 5, 4]

Solution

1. Step 1: Identify nodes visible from left at each level

   Level 0: 1, Level 1: 2 (left child), Level 2: 5 (right child of 2)

2. Step 2: Collect visible nodes in order

   Left side view nodes are [1, 2, 5]

3. Final Answer:

   [1, 2, 5] -> Option B

4. Quick Check:

   Left side nodes = [1, 2, 5] [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 Rust function snippet for left side view, what is the bug?

fn left_side_view(root: Option) -> Vec<i32> {
    let mut result = vec![];
    let mut queue = vec![root.unwrap()];
    while !queue.is_empty() {
        let level_size = queue.len();
        for i in 0..level_size {
            let node = queue.remove(0);
            if i == 0 {
                result.push(node.val);
            }
            if let Some(left) = node.left {
                queue.push(left);
            }
            if let Some(right) = node.right {
                queue.push(right);
            }
        }
    }
    result
}

medium

A. Calling unwrap() on root without checking if it is null

B. Using remove(0) on vector causes runtime error

C. Pushing right child before left child breaks left view order

D. Not returning the result vector at the end

Solution

1. Step 1: Check root unwrap usage

   Calling unwrap() on root without checking if it is null can cause panic if tree is empty.

2. Step 2: Confirm other parts are correct

   remove(0) works but is inefficient; order of pushing left then right is correct; result is returned.

3. Final Answer:

   Calling unwrap() on root without checking if it is null -> Option A

4. Quick Check:

   unwrap() on null causes panic [OK]

Hint: Check unwrap() calls for null before use [OK]

Common Mistakes:

• Ignoring unwrap panic possibility
• Thinking remove(0) causes error (just slow)
• Confusing child push order
• Missing return statement

5. You want to modify the left side view function to return the leftmost node values only for levels where the node value is even. Which approach is correct?

hard

A. Add a check inside the level loop: if i == 0 and node.val % 2 == 0, then push node.val to result

B. Push node.val for all nodes during traversal, then filter even values from result after traversal

C. Push node.val only if node.val is even, ignoring position in level

D. Push node.val if it is even and it is the rightmost node at the level

Solution

1. Step 1: Understand requirement for leftmost even nodes

   We want only the first node at each level if its value is even.

2. Step 2: Apply condition during traversal

   Check if i == 0 (leftmost) and node.val % 2 == 0 before pushing to result.

3. Final Answer:

   Add a check inside the level loop: if i == 0 and node.val % 2 == 0, then push node.val to result -> Option A

4. Quick Check:

   Filter leftmost nodes by even value during traversal [OK]

Hint: Check leftmost and even condition together inside loop [OK]

Common Mistakes:

• Filtering after traversal loses position info
• Ignoring leftmost position and filtering only by value
• Checking rightmost node instead of leftmost