Left Side View of Binary Tree in DSA Javascript - 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 Javascript

function leftSideView(root) {
  const result = [1];
  // rest of the code
}

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A0

B{}

Cnull

D[]

Attempts:

3 left

💡 Hint

Common Mistakes

Using an object {} instead of an array.

Initializing result as null or 0 which cannot hold multiple nodes.

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 visited for the first time.

DSA Javascript

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

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Anode.val

Blevel

Cresult.length

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing level with level itself instead of result length.

Using node.val in the condition which is incorrect.

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

3fill in blank

hard

Fix the error in the DFS traversal order to get the correct left side view.

DSA Javascript

function dfs(node, level) {
  if (!node) return;
  if (level === result.length) {
    result.push(node.val);
  }
  dfs(node.[1], level + 1);
  dfs(node.right, level + 1);
}

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Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Visiting right child before left child causes wrong nodes in left view.

Using invalid property names like parent or child.

We must visit the left child first to capture the leftmost nodes at each level.

4fill in blank

hard

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

DSA Javascript

function leftSideView(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  while (queue.length > 0) {
    const size = queue.length;
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (i === [1]) {
        result.push(node.val);
      }
      if (node.[2]) queue.push(node.[2]);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

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A0

B1

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using i === 1 instead of i === 0 to pick the first node.

Enqueuing right child before left child changes the order.

We add the first node at each level (i === 0) and enqueue left child first for left side view.

5fill in blank

hard

Fill all three blanks to complete the recursive left side view function with helper.

DSA Javascript

function leftSideView(root) {
  const result = [];
  function helper(node, level) {
    if (!node) return;
    if (level === [1]) {
      result.push(node.[2]);
    }
    helper(node.[3], level + 1);
    helper(node.right, level + 1);
  }
  helper(root, 0);
  return result;
}

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Aresult.length

Bval

Cleft

Dlevel

Attempts:

3 left

💡 Hint

Common Mistakes

Using level instead of result.length in condition.

Pushing node instead of node.val.

Recursing right child before left child.

We check if level equals result.length, push node.val, and recurse left child first.