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
}

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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);
}

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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);

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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;
}

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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;
}

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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.