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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to visit the root node first in preorder traversal.

DSA Javascript

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

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

A.left

B.right

C.parent

D.child

Attempts:

3 left

💡 Hint

Common Mistakes

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

Trying to access .parent instead of children.

In preorder traversal, after visiting the root, we visit the left child first.

2fill in blank

medium

Complete the code to visit the right child after the left child in preorder traversal.

DSA Javascript

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

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

A.right

B.left

C.parent

D.child

Attempts:

3 left

💡 Hint

Common Mistakes

Visiting .left twice instead of .right second time.

Trying to visit .parent or .child which are not correct.

After visiting the root and left subtree, preorder visits the right subtree.

3fill in blank

hard

Fix the error in the preorder traversal code to correctly visit root, left, then right.

DSA Javascript

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

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

A.left

B.right

C.parent

D.child

Attempts:

3 left

💡 Hint

Common Mistakes

Calling preorder on .left twice.

Using .parent or .child which are invalid here.

The second recursive call must visit the right child to follow preorder traversal order.

4fill in blank

hard

Fill both blanks to complete the preorder traversal visiting root, left, then right.

DSA Javascript

function preorder(node) {
  if (node == null) return;
  console.log(node[1]);
  preorder(node[2]);
  preorder(node.right);
}

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

A.value

B.left

C.right

D.parent

Attempts:

3 left

💡 Hint

Common Mistakes

Printing .left or .right instead of .value.

Visiting .right before .left.

First print the node's value, then recursively visit the left child, then the right child.

5fill in blank

hard

Fill all three blanks to complete preorder traversal: print root, visit left, then right.

DSA Javascript

function preorder(node) {
  if (node == null) return;
  console.log(node[1]);
  preorder(node[2]);
  preorder(node[3]);
}

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

A.value

B.left

C.right

D.parent

Attempts:

3 left

💡 Hint

Common Mistakes

Visiting right child before left child.

Printing .left or .right instead of .value.

Print the node's value, then visit left child, then right child for preorder traversal.

Practice

1. What is the order of nodes visited in a preorder tree traversal?

easy

A. Left, Right, Root

B. Left, Root, Right

C. Root, Left, Right

D. Right, Root, Left

Solution

1. Step 1: Understand preorder traversal definition

   Preorder traversal visits the root node first, then the left subtree, and finally the right subtree.

2. Step 2: Confirm the visiting order

   The visiting order is Root, then Left child, then Right child.

3. Final Answer:

   Root, Left, Right -> Option C

4. Quick Check:

   Preorder = Root, Left, Right [OK]

Hint: Remember preorder starts with the root node first [OK]

Common Mistakes:

• Confusing preorder with inorder or postorder
• Visiting left or right child before root
• Mixing up left and right subtree order

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

easy

A. function preorder(root) { if(root) { preorder(root.left); preorder(root.right); console.log(root.value); } }

B. function preorder(root) { if(root) { preorder(root.left); console.log(root.value); preorder(root.right); } }

C. function preorder(root) { if(root) { preorder(root.right); preorder(root.left); console.log(root.value); } }

D. function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } }

Solution

1. Step 1: Recall preorder traversal order

   Preorder visits root first, then left subtree, then right subtree.

2. Step 2: Check function steps

   function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } } prints root value first, then recursively visits left and right children, matching preorder.

3. Final Answer:

   function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } } -> Option D

4. Quick Check:

   Print root before children = function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } } [OK]

Hint: Print root value before recursive calls for preorder [OK]

Common Mistakes:

• Printing after visiting children (postorder)
• Swapping left and right recursive calls
• Missing base case for null root

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

    1
   / \
  2   3
 / \
4   5

Use the format: 1 -> 2 -> 4 -> 5 -> 3 -> null

medium

A. 1 -> 2 -> 4 -> 5 -> 3 -> null

B. 4 -> 2 -> 5 -> 1 -> 3 -> null

C. 2 -> 4 -> 5 -> 1 -> 3 -> null

D. 1 -> 3 -> 2 -> 4 -> 5 -> null

Solution

1. Step 1: Start at root node 1

   Visit root first: 1

2. Step 2: Traverse left subtree of 1

   Visit 2, then its left child 4, then right child 5

3. Step 3: Traverse right subtree of 1

   Visit 3

4. Final Answer:

   1 -> 2 -> 4 -> 5 -> 3 -> null -> Option A

5. Quick Check:

   Preorder visits root, left, right = 1 -> 2 -> 4 -> 5 -> 3 -> null [OK]

Hint: Follow root, then left subtree, then right subtree order [OK]

Common Mistakes:

• Visiting left subtree after right subtree
• Starting traversal from left child instead of root
• Confusing preorder with inorder or postorder

4. The following JavaScript code is intended to perform a preorder traversal and print node values. What is the error?

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

medium

A. The console.log should be before the recursive calls

B. The base case should check for undefined, not null

C. The function should return a value instead of void

D. The left and right recursive calls are swapped

Solution

1. Step 1: Analyze the order of operations

   The code visits left child first, then prints root, then visits right child.

2. Step 2: Compare with preorder traversal order

   Preorder requires printing root before visiting children, so console.log should be before recursive calls.

3. Final Answer:

   The console.log should be before the recursive calls -> Option A

4. Quick Check:

   Print root first in preorder [OK]

Hint: Print root value before recursive calls in preorder [OK]

Common Mistakes:

• Printing after left subtree instead of before
• Confusing preorder with inorder traversal
• Ignoring base case for null nodes

5. You have a binary tree where some nodes have only one child. How does preorder traversal handle nodes with a missing left child?
Consider this tree:

    10
      \
      20
     /  \
   30    40

What is the preorder traversal output?

hard

A. 10 -> 40 -> 20 -> 30 -> null

B. 10 -> 20 -> 30 -> 40 -> null

C. 10 -> 20 -> 40 -> 30 -> null

D. 10 -> 30 -> 20 -> 40 -> null

Solution

1. Step 1: Visit root node 10

   Start with root: 10

2. Step 2: Check left child of 10

   Left child is missing, so skip to right child 20

3. Step 3: Visit node 20 and its children

   Visit 20, then left child 30, then right child 40

4. Final Answer:

   10 -> 20 -> 30 -> 40 -> null -> Option B

5. Quick Check:

   Preorder visits root, left (if any), right = 10 -> 20 -> 30 -> 40 -> null [OK]

Hint: Skip missing children, visit existing nodes in preorder order [OK]

Common Mistakes:

• Assuming missing left child means skip right child
• Visiting children in wrong order
• Confusing preorder with other traversals