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

Time Complexity: Left Side View of Binary Tree

O(n)

Understanding Time Complexity

We want to understand 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 to get the left side view?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the left side view by visiting each level of the tree and recording the first node at each level.

Identify Repeating Operations

• Primary operation: The while loop runs once per level, and the inner for loop visits every node in the tree exactly once.
• How many times: Each node is processed once, so the total number of operations grows with the number of nodes n.

How Execution Grows With Input

As the number of nodes increases, the code visits each node once, so the work grows directly with the number of nodes.

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

Pattern observation: The operations increase linearly as the tree size grows.

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 take one node per level, the time complexity is O(height) or O(log n)."

[OK] Correct: Even though we record one node per level, we still visit every node to check their children, so the total work depends on all nodes, not just the levels.

Interview Connect

Understanding this helps you explain how breadth-first search visits nodes level by level, a common pattern in tree problems.

Self-Check

What if we changed the code to use recursion instead of a queue? 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 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