Left Side View of Binary Tree in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the result array for storing the left side view nodes.

DSA Typescript

function leftSideView(root: TreeNode | null): number[] {
  const result = [1];
  // rest of the code
}

Drag options to blanks, or click blank then click option'

A[]

B{}

Cnull

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using an object {} instead of an array.

Initializing with null or 0 which are not containers.

We use an empty array [] to store the nodes visible from the left side.

2fill in blank

medium

Complete the code to check if the current level is the first time we visit it.

DSA Typescript

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

Drag options to blanks, or click blank then click option'

Aresult.length

Blevel

C0

Dnode.val

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing level with level variable itself.

Using node.val instead of result length.

We compare the current level with the length of result to add the first node at that level.

3fill in blank

hard

Fix the error in the recursive calls to traverse the tree correctly for left side view.

DSA Typescript

dfs(node.[1], level + 1);
dfs(node.[2], level + 1);

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cchild

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right child traversal order.

Using invalid properties like child or parent.

We first traverse the left child, then the right child to get the left side view.

4fill in blank

hard

Fill both blanks to complete the function that returns the left side view of the binary tree.

DSA Typescript

function leftSideView(root: TreeNode | null): number[] {
  const result = [];
  function dfs(node: TreeNode | null, level: number) {
    if (!node) return;
    if (level === [1]) {
      result.push(node.val);
    }
    dfs(node.[2], level + 1);
    dfs(node.right, level + 1);
  }
  dfs(root, 0);
  return result;
}

Drag options to blanks, or click blank then click option'

Aresult.length

Blevel

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using level instead of result.length in condition.

Traversing right child before left child.

We check if level equals result.length and traverse left child first.

5fill in blank

hard

Fill all three blanks to complete the iterative approach using a queue for left side view.

DSA Typescript

function leftSideView(root: TreeNode | null): number[] {
  if (!root) return [];
  const result: number[] = [];
  const queue: TreeNode[] = [root];
  while (queue.length > 0) {
    const levelSize = queue.length;
    for (let i = 0; i < levelSize; i++) {
      const node = queue.shift()!;
      if (i === [1]) {
        result.push(node.[2]);
      }
      if (node.[3]) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

Drag options to blanks, or click blank then click option'

A0

Bval

Cleft

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i === 1 instead of i === 0 for first node.

Accessing wrong property instead of val.

Enqueueing right child before left child.

We add the first node at each level (i === 0), push its value, and enqueue left child first.

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