Tree Traversal Inorder Left Root Right in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to visit the left subtree first in inorder traversal.

DSA Javascript

function inorder(node) {
  if (node === null) return;
  inorder(node.[1]);
  console.log(node.value);
  inorder(node.right);
}

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

Aparent

Bright

Croot

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' causes wrong traversal order.

Using 'root' or 'parent' properties which do not exist.

In inorder traversal, we first visit the left child of the current node.

2fill in blank

medium

Complete the code to print the current node's value in inorder traversal.

DSA Javascript

function inorder(node) {
  if (node === null) return;
  inorder(node.left);
  console.log(node.[1]);
  inorder(node.right);
}

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

Avalue

Bleft

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Printing 'left' or 'right' instead of the node's value.

Trying to print 'parent' which is not defined.

We print the current node's value after visiting the left subtree.

3fill in blank

hard

Fix the error in the inorder traversal to visit the right subtree correctly.

DSA Javascript

function inorder(node) {
  if (node === null) return;
  inorder(node.left);
  console.log(node.value);
  inorder(node.[1]);
}

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

Aroot

Bleft

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' causes wrong traversal.

Using 'root' or 'parent' which are invalid here.

After visiting the root, inorder traversal visits the right child.

4fill in blank

hard

Fill both blanks to create a function that returns an array of inorder traversal values.

DSA Javascript

function inorderArray(node) {
  if (node === null) return [];
  return [...inorderArray(node.[1]), node.[2], ...inorderArray(node.right)];
}

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

Aleft

Bvalue

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right causes wrong order.

Using 'parent' or 'root' which are invalid.

We collect values from left subtree, then current node's value, then right subtree.

5fill in blank

hard

Fill all three blanks to create a function that builds a map of node values to their inorder positions.

DSA Javascript

function inorderMap(node, map = new Map(), index = {count: 0}) {
  if (node === null) return map;
  inorderMap(node.[1], map, index);
  map.set(node.[2], index.[3]);
  index.count++;
  inorderMap(node.right, map, index);
  return map;
}

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

Aleft

Bvalue

Ccount

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' in first blank.

Using 'parent' or 'root' which do not exist.

Using 'index' instead of 'count' for the counter.

We traverse left subtree, map node's value to current count, then increment count and traverse right subtree.

Practice

1. What is the order of nodes visited in an inorder traversal of a binary tree?

easy

A. Root, Right subtree, Left subtree

B. Left subtree, Root, Right subtree

C. Right subtree, Root, Left subtree

D. Root, Left subtree, Right subtree

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal visits nodes in the order: left subtree first, then the root node, and finally the right subtree.

2. Step 2: Match the traversal order to options

   Left subtree, Root, Right subtree correctly states the order as Left subtree, Root, Right subtree.

3. Final Answer:

   Left subtree, Root, Right subtree -> Option B

4. Quick Check:

   Inorder = Left, Root, Right [OK]

Hint: Inorder always visits left child before root [OK]

Common Mistakes:

• Confusing inorder with preorder or postorder
• Visiting root before left subtree
• Visiting right subtree before root

2. Which of the following is the correct JavaScript function signature for inorder traversal of a binary tree node named node?

easy

A. function inorder(node) { inorder(node.right); console.log(node.value); inorder(node.left); }

B. function inorder(node) { console.log(node.value); inorder(node.left); inorder(node.right); }

C. function inorder(node) { if(node) { inorder(node.left); console.log(node.value); inorder(node.right); } }

D. function inorder(node) { inorder(node.left); inorder(node.right); console.log(node.value); }

Solution

1. Step 1: Recall inorder traversal steps in code

   Inorder traversal visits left subtree, then root, then right subtree. So the function must call inorder on left, print root, then call inorder on right.

2. Step 2: Check each option's order

   function inorder(node) { if(node) { inorder(node.left); console.log(node.value); inorder(node.right); } } matches the correct order: left, root (console.log), right. Others have wrong orders.

3. Final Answer:

   function inorder(node) { if(node) { inorder(node.left); console.log(node.value); inorder(node.right); } } -> Option C

4. Quick Check:

   Left, Root, Right in code = function inorder(node) { if(node) { inorder(node.left); console.log(node.value); inorder(node.right); } } [OK]

Hint: Inorder prints after left call, before right call [OK]

Common Mistakes:

• Printing before left subtree traversal
• Swapping left and right subtree calls
• Missing null check for node

3. Given the binary tree below, what is the output of inorder traversal?

    10
   /  \
  5    15
 /    /  \
3    12  20

medium

A. [3, 5, 10, 12, 15, 20]

B. [10, 5, 3, 15, 12, 20]

C. [3, 5, 12, 10, 15, 20]

D. [5, 3, 10, 15, 12, 20]

Solution

1. Step 1: Traverse left subtree of root 10

   Left subtree of 10 is rooted at 5. Its left child is 3. Inorder of left subtree: visit 3, then 5 (root of subtree).

2. Step 2: Visit root 10 and then right subtree

   After left subtree, visit root 10. Then traverse right subtree rooted at 15: left child 12, root 15, right child 20. So inorder right subtree is 12, 15, 20.

3. Final Answer:

   [3, 5, 10, 12, 15, 20] -> Option A

4. Quick Check:

   Inorder = Left, Root, Right = [3,5,10,12,15,20] [OK]

Hint: List nodes left-root-right recursively [OK]

Common Mistakes:

• Visiting root before left subtree
• Mixing order of right subtree nodes
• Skipping leaf nodes

4. The following JavaScript code is intended to perform inorder traversal but has a bug. What is the bug?

function inorder(node) {
  if (node === null) return;
  console.log(node.value);
  inorder(node.left);
  inorder(node.right);
}

medium

A. It prints the root before left subtree, so not inorder

B. It does not check for null nodes

C. It misses traversing the right subtree

D. It calls inorder on left and right in wrong order

Solution

1. Step 1: Analyze the order of operations in code

   The code prints node.value first, then calls inorder on left and right. This is preorder traversal order.

2. Step 2: Identify correct inorder order

   Inorder traversal requires visiting left subtree first, then printing root, then right subtree. The code prints root before left subtree, which is incorrect.

3. Final Answer:

   It prints the root before left subtree, so not inorder -> Option A

4. Quick Check:

   Print after left call for inorder [OK]

Hint: Print after left subtree call for inorder [OK]

Common Mistakes:

• Printing root before left subtree
• Confusing preorder with inorder
• Not checking for null nodes (though code does here)

5. You have a binary search tree (BST) and want to get all values in ascending order using inorder traversal. Which of the following modifications to the inorder traversal function will correctly collect values into an array instead of printing them?

function inorder(node, arr) {
  if (!node) return;
  // Fill missing code here
}

hard

A. inorder(node.left, arr); inorder(node.right, arr); arr.push(node.value);

B. arr.push(node.value); inorder(node.left, arr); inorder(node.right, arr);

C. inorder(node.right, arr); arr.push(node.value); inorder(node.left, arr);

D. inorder(node.left, arr); arr.push(node.value); inorder(node.right, arr);

Solution

1. Step 1: Recall inorder traversal order for BST

   Inorder traversal visits left subtree, then root, then right subtree, which returns values in ascending order for BST.

2. Step 2: Choose code that collects values in correct order

   inorder(node.left, arr); arr.push(node.value); inorder(node.right, arr); calls inorder on left, then pushes current node value, then calls inorder on right, correctly collecting values in ascending order.

3. Final Answer:

   inorder(node.left, arr); arr.push(node.value); inorder(node.right, arr); -> Option D

4. Quick Check:

   Left, Root, Right with push = inorder(node.left, arr); arr.push(node.value); inorder(node.right, arr); [OK]

Hint: Push after left call, before right call for sorted array [OK]

Common Mistakes:

• Pushing before left subtree traversal
• Swapping left and right subtree calls
• Pushing after both subtree traversals