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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

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

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

A.left

B.val

C.parent

D.right

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.right instead of node.left for the first recursive call.

Trying to access node.val instead of a child node.

In inorder traversal, we first visit the left child, so we call inorder(node.left).

2fill in blank

medium

Complete the code to print the root node value in inorder traversal.

DSA Typescript

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

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

A.parent

B.right

C.left

D.val

Attempts:

3 left

💡 Hint

Common Mistakes

Printing node.left or node.right instead of node.val.

Trying to print node.parent which may not exist.

In inorder traversal, after visiting left subtree, we print the root node's value using node.val.

3fill in blank

hard

Fix the error in the inorder traversal recursive call for the right subtree.

DSA Typescript

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

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

A.parent

B.left

C.right

D.val

Attempts:

3 left

💡 Hint

Common Mistakes

Calling inorder(node.left) again instead of node.right.

Trying to call inorder(node.val) which is a value, not a node.

The right subtree is visited by calling inorder(node.right) in inorder traversal.

4fill in blank

hard

Fill both blanks to complete the inorder traversal that collects values in an array.

DSA Typescript

function inorder(node: TreeNode | null, result: number[]): void {
  if (node === null) return;
  inorder(node[1], result);
  result.push(node[2]);
  inorder(node.right, result);
}

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

A.left

B.val

C.right

D.parent

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.right in the first recursive call instead of node.left.

Pushing node.right or node.left instead of node.val.

We first recurse on the left child (node.left), then push the root value (node.val) to the result array.

5fill in blank

hard

Fill all three blanks to complete inorder traversal that returns an array of values.

DSA Typescript

function inorder(node: TreeNode | null): number[] {
  if (node === null) return [];
  const left = inorder(node[1]);
  const right = inorder(node[2]);
  return [...left, node[3], ...right];
}

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

A.left

B.right

C.val

D.parent

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right children in recursive calls.

Using node.parent or node.val in recursive calls instead of child nodes.

Returning node instead of node.val in the combined array.

We recursively get left subtree values with node.left, right subtree values with node.right, and combine them with the root value node.val in the middle.

Practice

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

easy

A. Left subtree, Root, Right subtree

B. Root, Left subtree, Right subtree

C. Right subtree, Root, Left subtree

D. Root, Right subtree, Left 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 A

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 TypeScript function signatures correctly represents an inorder traversal function for a binary tree node?

easy

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

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

C. function inorder(): void

D. function inorder(node: TreeNode | null): number

Solution

1. Step 1: Analyze function parameters and return type

   An inorder traversal function typically takes a node or null as input and performs traversal, often printing or processing nodes without returning a value.

2. Step 2: Check options for correct signature

   function inorder(node: TreeNode | null): void accepts a node or null and returns void, which fits traversal that prints or processes nodes.

3. Final Answer:

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

4. Quick Check:

   Traversal function takes node or null, returns void [OK]

Hint: Traversal functions usually take node or null, return void [OK]

Common Mistakes:

• Omitting null in parameter type
• Returning incorrect types like number or array
• Missing parameter in function signature

3. Given the following binary tree:

    2
   / \
  1   3

What is the output of inorder traversal printing node values?

medium

A. [2, 1, 3]

B. [3, 2, 1]

C. [1, 3, 2]

D. [1, 2, 3]

Solution

1. Step 1: Visualize the tree and inorder traversal

   Inorder visits left subtree, root, then right subtree. For root 2, left child is 1, right child is 3.

2. Step 2: List nodes in inorder

   Visit left: 1, then root: 2, then right: 3, so output is [1, 2, 3].

3. Final Answer:

   [1, 2, 3] -> Option D

4. Quick Check:

   Inorder of 2-node tree = [1, 2, 3] [OK]

Hint: Inorder prints nodes sorted for BST [OK]

Common Mistakes:

• Printing root before left subtree
• Swapping left and right subtree order
• Confusing preorder or postorder output

4. Consider this TypeScript inorder traversal function:

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

What is the main problem with this code?

medium

A. It prints node values twice

B. It does not handle null nodes properly

C. It visits right subtree before left, so traversal order is incorrect

D. It causes infinite recursion

Solution

1. Step 1: Analyze traversal order in code

   The function calls inorder on right subtree first, then prints root, then calls inorder on left subtree.

2. Step 2: Compare with correct inorder order

   Correct inorder is left subtree, root, right subtree. This code reverses left and right, so order is wrong.

3. Final Answer:

   It visits right subtree before left, so traversal order is incorrect -> Option C

4. Quick Check:

   Inorder must visit left before right [OK]

Hint: Inorder must call left subtree before right [OK]

Common Mistakes:

• Swapping left and right subtree calls
• Forgetting base case for null
• Printing before recursive calls

5. You have a binary tree where each node contains a number. You want to collect all node values in an array using inorder traversal. Which TypeScript code snippet correctly implements this?

hard

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

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

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

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

Solution

1. Step 1: Understand inorder traversal with array accumulation

   We must visit left subtree, then root (push value), then right subtree, passing the same array to collect values.

2. Step 2: Check each option for correct order and array usage

   function inorder(node: TreeNode | null, arr: number[]): void { if (!node) return; inorder(node.left, arr); arr.push(node.val); inorder(node.right, arr); } correctly visits left, pushes root value, then visits right, using the same array.

3. Final Answer:

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

4. Quick Check:

   Left, root, right with shared array [OK]

Hint: Push node value after left subtree call [OK]

Common Mistakes:

• Pushing before left subtree traversal
• Swapping left and right subtree calls
• Returning new arrays instead of using passed array