Left Side View of Binary Tree in DSA Typescript - Execution Trace

Concept Flow - Left Side View of Binary Tree

Start at root node

↓

Initialize queue with root

↓

While queue not empty

↓

Get number of nodes at current level

↓

For each node at this level

↓

Dequeue node

↓

Enqueue left child if exists

↓

Enqueue right child if exists

↓

Repeat for next level

↓

Return collected left side view nodes

We start from the root and use a queue to traverse the tree level by level. For each level, we record the first node encountered as part of the left side view.

Execution Sample

DSA Typescript

function leftSideView(root: TreeNode | null): number[] {
  if (!root) return [];
  const queue = [root];
  const leftView = [];
  while (queue.length) {
    const levelSize = queue.length;
    for (let i = 0; i < levelSize; i++) {
      const node = queue.shift()!;
      if (i === 0) leftView.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return leftView;
}

This code collects the left side view nodes of a binary tree using level order traversal.

Execution Table

Step	Operation	Node Visited	Queue State	Left View Collected	Visual State
1	Initialize queue with root	1	[1]	[]	1 / \ 2 3 / \ 4 5
2	Start level traversal, levelSize=1	1	[1]	[]	1 / \ 2 3 / \ 4 5
3	Visit first node at level, add to left view	1	[]	[1]	1 / \ 2 3 / \ 4 5
4	Enqueue left child of 1	1	[2]	[1]	1 / \ 2 3 / \ 4 5
5	Enqueue right child of 1	1	[2,3]	[1]	1 / \ 2 3 / \ 4 5
6	Start level traversal, levelSize=2	2	[2,3]	[1]	1 / \ 2 3 / \ 4 5
7	Visit first node at level, add to left view	2	[3]	[1,2]	1 / \ 2 3 / \ 4 5
8	Enqueue left child of 2	2	[3,4]	[1,2]	1 / \ 2 3 / \ 4 5
9	No right child of 2	2	[3,4]	[1,2]	1 / \ 2 3 / \ 4 5
10	Visit next node at level	3	[4]	[1,2]	1 / \ 2 3 / \ 4 5
11	No left child of 3	3	[4]	[1,2]	1 / \ 2 3 / \ 4 5
12	Enqueue right child of 3	3	[4,5]	[1,2]	1 / \ 2 3 / \ 4 5
13	Start level traversal, levelSize=2	4	[4,5]	[1,2]	1 / \ 2 3 / \ 4 5
14	Visit first node at level, add to left view	4	[5]	[1,2,4]	1 / \ 2 3 / \ 4 5
15	No left child of 4	4	[5]	[1,2,4]	1 / \ 2 3 / \ 4 5
16	No right child of 4	4	[5]	[1,2,4]	1 / \ 2 3 / \ 4 5
17	Visit next node at level	5	[]	[1,2,4]	1 / \ 2 3 / \ 4 5
18	No left child of 5	5	[]	[1,2,4]	1 / \ 2 3 / \ 4 5
19	No right child of 5	5	[]	[1,2,4]	1 / \ 2 3 / \ 4 5
20	Queue empty, traversal ends	-	[]	[1,2,4]	1 / \ 2 3 / \ 4 5

💡 Queue is empty, all levels processed, left side view collected

Variable Tracker

Variable	Start	After Step 4	After Step 6	After Step 13	Final
queue	[1]	[2]	[2,3]	[4,5]	[]
leftView	[]	[1]	[1]	[1,2]	[1,2,4]
levelSize	-	1	2	2	-

Key Moments - 3 Insights

Why do we add only the first node of each level to the left view?

Because the first node visited at each level is the leftmost node visible from the left side. This is shown in execution_table rows 3, 7, and 14 where the first node at each level is added to leftView.

Why do we use a queue for traversal?

A queue helps us visit nodes level by level (breadth-first). This ensures we process all nodes at one level before moving to the next, as seen in the queue state changes in execution_table.

What happens if the tree is empty?

The function returns an empty array immediately (see execution_sample code line 2). No traversal happens and leftView remains empty.

Visual Quiz - 3 Questions

Look at the execution_table, what is the leftView collected after visiting node 2 at step 7?

A []

B [1]

C [1,2]

D [1,2,4]

💡 Hint

Check the 'Left View Collected' column at step 7 in execution_table

At which step does the queue become empty, ending the traversal?

A Step 17

B Step 20

C Step 19

D Step 13

💡 Hint

Look for the step where 'Queue State' is empty and operation notes traversal ends

If node 2 had a right child, how would the queue state change after step 8?

A Queue would include the right child after node 4

B Queue would remain the same as [3,4]

C Queue would only have the right child

D Queue would be empty

💡 Hint

Refer to steps 4 and 5 where children are enqueued after visiting a node

Concept Snapshot

Left Side View of Binary Tree:
- Use level order traversal with a queue
- For each level, record the first node visited
- Enqueue left then right children
- Continue until queue is empty
- Result is nodes visible from left side

Full Transcript

This visualization shows how to find the left side view of a binary tree. We start at the root and use a queue to visit nodes level by level. At each level, the first node visited is added to the left view list. We enqueue children left first, then right, to ensure leftmost nodes come first. The process repeats until all nodes are visited and the queue is empty. The final left view contains the nodes visible from the left 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 right side

B. Nodes at the deepest level of the tree

C. All leaf nodes of the tree

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

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 D

4. Quick Check:

   Left side view = nodes visible from left [OK]

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

Common Mistakes:

• Confusing left side view with right side view
• Thinking it shows all leaf nodes
• Assuming it shows deepest nodes only

2. Which of the following TypeScript function signatures correctly represents a function to get the left side view of a binary tree?

easy

A. function leftSideView(root: number[]): number

B. function leftSideView(root: TreeNode | null): number[]

C. function leftSideView(root: TreeNode): void

D. function leftSideView(root: string): string[]

Solution

1. Step 1: Identify correct input type

   The function should accept a binary tree root node or null, so TreeNode | null is correct.

2. Step 2: Identify correct return type

   The function returns an array of numbers representing node values visible from the left side.

3. Final Answer:

   function leftSideView(root: TreeNode | null): number[] -> Option B

4. Quick Check:

   Input = TreeNode or null, Output = number array [OK]

Hint: Input is tree root or null; output is array of numbers [OK]

Common Mistakes:

• Using array of numbers as input instead of tree node
• Returning void instead of array
• Using wrong input type like string

3. Given the binary tree below, what is the left side view output?

    1
   / \
  2   3
   \   \
    4   5

Consider the TypeScript code:
console.log(leftSideView(root));

medium

A. [1, 2, 4]

B. [1, 3, 5]

C. [1, 2, 5]

D. [1, 4, 5]

Solution

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

   Level 1: Node 1 is visible.
   Level 2: Node 2 is the leftmost node.
   Level 3: Node 4 is visible from left (child of 2).

2. Step 2: Collect visible nodes in order

   The left side view nodes are [1, 2, 4].

3. Final Answer:

   [1, 2, 4] -> Option A

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
• Confusing order of nodes

4. The following TypeScript code attempts to find the left side view of a binary tree but has a bug. What is the bug?

function leftSideView(root: TreeNode | null): number[] {
  const result: number[] = [];
  function dfs(node: TreeNode | null, level: number) {
    if (!node) return;
    if (result.length === level) {
      result.push(node.val);
    }
    dfs(node.right, level + 1);
    dfs(node.left, level + 1);
  }
  dfs(root, 0);
  return result;
}

medium

A. The result array is not initialized

B. The base case is missing to stop recursion

C. The function visits right child before left, so left side nodes are missed

D. The function does not return any value

Solution

1. Step 1: Analyze traversal order

   The code visits the right child before the left child, so it records right side nodes first.

2. Step 2: Understand impact on left side view

   Because right nodes are visited first, left side nodes at each level may be skipped, causing incorrect output.

3. Final Answer:

   The function visits right child before left, so left side nodes are missed -> Option C

4. Quick Check:

   Visit left child first for correct left side view [OK]

Hint: Traverse left child before right to capture left side nodes [OK]

Common Mistakes:

• Visiting right child before left child
• Forgetting base case (though present here)
• Not pushing node values at correct level

5. You want to find the left side view of a binary tree but only have the tree nodes stored in a nested object format without parent references. Which approach is best to implement in TypeScript to get the left side view efficiently?

hard

A. Use level order traversal (BFS) and pick the first node at each level

B. Use in-order traversal and collect nodes in an array

C. Use post-order traversal and reverse the result array

D. Use a stack to perform iterative right-to-left traversal

Solution

1. Step 1: Understand traversal types and their suitability

   Level order traversal (BFS) visits nodes level by level, making it easy to pick the first node at each level.

2. Step 2: Evaluate other traversal methods

   In-order and post-order traversals do not guarantee level-wise order. Right-to-left iterative traversal misses left side nodes.

3. Final Answer:

   Use level order traversal (BFS) and pick the first node at each level -> Option A

4. Quick Check:

   BFS level order = best for left side view [OK]

Hint: Use BFS and select first node per level for left side view [OK]

Common Mistakes:

• Using in-order or post-order which don't give level order
• Traversing right to left instead of left to right
• Ignoring level boundaries in traversal