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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start preorder traversal by visiting the root node first.

DSA Typescript

function preorder(node: TreeNode | null): void {
  if (node === null) return;
  console.log(node.val);
  preorder([1]);
}

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

Anode

Bnode.right

Cnode.left

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Calling preorder on right child first

Calling preorder on the node itself again causing infinite recursion

In preorder traversal, after visiting the root, we visit the left subtree first, so we call preorder on node.left.

2fill in blank

medium

Complete the code to continue preorder traversal by visiting the right subtree after the left.

DSA Typescript

function preorder(node: TreeNode | null): void {
  if (node === null) return;
  console.log(node.val);
  preorder(node.left);
  preorder([1]);
}

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

Anode.right

Bnode.left

Cnode

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Calling preorder on left child twice

Calling preorder on the node itself again

After visiting the root and left subtree, preorder traversal visits the right subtree, so we call preorder on node.right.

3fill in blank

hard

Fix the error in the preorder traversal function to avoid infinite recursion.

DSA Typescript

function preorder(node: TreeNode | null): void {
  if (node === null) return;
  console.log(node.val);
  preorder(node.left);
  preorder([1]);
}

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

Anode

Bnode.right

Cnull

Dnode.left

Attempts:

3 left

💡 Hint

Common Mistakes

Calling preorder on node itself

Calling preorder on null unnecessarily

Calling preorder on node again causes infinite recursion. We must call preorder on node.right to continue traversal correctly.

4fill in blank

hard

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

DSA Typescript

function preorder(node: TreeNode | null): void {
  if (node === null) return;
  console.log(node.val);
  preorder([1]);
  preorder([2]);
}

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

Anode.left

Bnode.right

Cnode

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right calls

Calling preorder on the node itself

Preorder traversal visits root first, then recursively visits left subtree (node.left), then right subtree (node.right).

5fill in blank

hard

Fill all three blanks to create a preorder traversal that prints the node value in uppercase, then visits left and right children.

DSA Typescript

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

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

A.toUpperCase()

Bnode.left

Cnode.right

D.toLowerCase()

Attempts:

3 left

💡 Hint

Common Mistakes

Using .toLowerCase() instead of .toUpperCase()

Swapping left and right children calls

To print the node value in uppercase, use .toUpperCase(). Then preorder visits left child (node.left) and right child (node.right).

Practice

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

easy

A. Root, Left, Right

B. Left, Root, Right

C. Left, Right, Root

D. Right, Left, Root

Solution

1. Step 1: Understand preorder traversal definition

   Preorder traversal visits the root node first, then recursively visits the left subtree, followed by the right subtree.

2. Step 2: Match the order with options

   The order Root, Left, Right matches the preorder traversal definition exactly.

3. Final Answer:

   Root, Left, Right -> Option A

4. Quick Check:

   Preorder = Root, Left, Right [OK]

Hint: Preorder always starts with the root node first [OK]

Common Mistakes:

• Confusing preorder with inorder or postorder
• Visiting left or right before root
• Mixing up traversal orders

2. Which of the following TypeScript function signatures correctly represents a preorder traversal function for a binary tree node?

easy

A. function preorder(node: TreeNode): number[]

B. function preorder(node: TreeNode | null): void

C. function preorder(): TreeNode

D. function preorder(node: TreeNode, result: number[]): number

Solution

1. Step 1: Analyze function parameters and return type

   A preorder traversal function typically takes a node (or null) and returns void if it prints or processes nodes directly.

2. Step 2: Check each option

   function preorder(node: TreeNode | null): void accepts a node or null and returns void, which fits a traversal that processes nodes without returning a value.

3. Final Answer:

   function preorder(node: TreeNode | null): void -> Option B

4. Quick Check:

   Traversal function signature with node or null and void return [OK]

Hint: Traversal functions usually take a node and return void or array [OK]

Common Mistakes:

• Missing null in node parameter type
• Wrong return type for traversal
• Incorrect number of parameters

3. Given the following binary tree:

    1
   / \
  2   3
 / \
4   5

What is the output of a preorder traversal (Root, Left, Right)?

medium

A. [4, 2, 5, 1, 3]

B. [2, 4, 5, 1, 3]

C. [1, 2, 4, 5, 3]

D. [1, 3, 2, 4, 5]

Solution

1. Step 1: Start at root node 1

   Preorder visits root first: 1

2. Step 2: Traverse left subtree of 1

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

3. Step 3: Traverse right subtree of 1

   Visit node 3: 3

4. Final Answer:

   [1, 2, 4, 5, 3] -> Option C

5. Quick Check:

   Preorder = Root, Left, Right = [1, 2, 4, 5, 3] [OK]

Hint: Visit root first, then left subtree, then right subtree [OK]

Common Mistakes:

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

4. Consider this TypeScript preorder traversal function:

function preorder(node: TreeNode | null): void {
  if (node === null) return;
  preorder(node.left);
  console.log(node.val);
  preorder(node.right);
}

What is the error in this code?

medium

A. It does not handle null nodes

B. It will cause infinite recursion

C. It prints nodes in postorder

D. It visits nodes in inorder, not preorder

Solution

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

   The function first calls preorder on left child, then prints current node, then calls preorder on right child.

2. Step 2: Compare with preorder definition

   Preorder should print the node first, then traverse left and right. This code prints after left subtree, so it is inorder traversal.

3. Final Answer:

   It visits nodes in inorder, not preorder -> Option D

4. Quick Check:

   Print before recursive calls = preorder; here print is in middle [OK]

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

Common Mistakes:

• Printing node after left subtree instead of before
• Confusing traversal orders
• Missing base case for null

5. You want to collect all node values of a binary tree in preorder traversal into an array. Which TypeScript function correctly implements this?

hard

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

B. function preorder(node: TreeNode | null): number[] {
if (!node) return [];
return [...preorder(node.left), node.val, ...preorder(node.right)];
}

C. function preorder(node: TreeNode | null): number[] {
if (!node) return [];
return [...preorder(node.left), ...preorder(node.right), node.val];
}

D. function preorder(node: TreeNode | null): number[] {
if (!node) return [];
return [node.val, ...preorder(node.right), ...preorder(node.left)];
}

Solution

1. Step 1: Understand preorder traversal collection

   Preorder visits root first, then left subtree, then right subtree, collecting values in order.

2. Step 2: Check each function's order and correctness

   function preorder(node: TreeNode | null, result: number[] = []): number[] { if (!node) return result; result.push(node.val); preorder(node.left, result); preorder(node.right, result); return result; } pushes node.val first, then recursively collects left and right, returning the result array correctly.

3. Final Answer:

   function preorder(node: TreeNode | null, result: number[] = []): number[] { if (!node) return result; result.push(node.val); preorder(node.left, result); preorder(node.right, result); return result; } -> Option A

4. Quick Check:

   Push root first, then left and right recursively [OK]

Hint: Push root value before recursive calls to left and right [OK]

Common Mistakes:

• Collecting left subtree before root
• Mixing left and right subtree order
• Returning new arrays instead of accumulating