Tree Traversal Inorder Left Root Right in DSA Typescript - Build from Scratch

Tree Traversal Inorder Left Root Right

📖 Scenario: You have a simple family tree where each person can have a left child and a right child. You want to visit each person in a special order: first the left child, then the person, then the right child. This is called inorder traversal.

🎯 Goal: Build a small tree using nodes, then write code to visit and print the nodes in inorder (left, root, right) order.

📋 What You'll Learn

Create a TreeNode class with value, left, and right properties

Create a tree with exactly 3 nodes: root with value 1, left child with value 2, right child with value 3

Write a function called inorderTraversal that visits nodes in left-root-right order

Print the values of nodes in inorder traversal separated by spaces

💡 Why This Matters

🌍 Real World

Tree traversal is used in many real-world tasks like searching family trees, organizing files, or evaluating expressions.

💼 Career

Understanding tree traversal is important for software developers working with data structures, databases, and algorithms.

1

Create TreeNode class and build the tree

Create a class called TreeNode with a constructor that takes a value (number) and sets left and right to null. Then create three nodes: root with value 1, root.left with value 2, and root.right with value 3.

DSA Typescript

# Your code here

Hint

Define the class with value, left, and right. Then create the root and assign left and right children.

2

Create an array to store traversal results

Create an empty array called result to store the values of nodes visited during inorder traversal.

DSA Typescript

class TreeNode {
    value: number;
    left: TreeNode | null;
    right: TreeNode | null;

    constructor(value: number) {
        this.value = value;
        this.left = null;
        this.right = null;
    }
}

const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);

# Your code here

Hint

Use const result: number[] = []; to create an empty array for storing values.

3

Write the inorder traversal function

Write a function called inorderTraversal that takes a node of type TreeNode | null. If the node is not null, first call inorderTraversal on node.left, then push node.value to result, then call inorderTraversal on node.right.

DSA Typescript

class TreeNode {
    value: number;
    left: TreeNode | null;
    right: TreeNode | null;

    constructor(value: number) {
        this.value = value;
        this.left = null;
        this.right = null;
    }
}

const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);

const result: number[] = [];

# Your code here

Hint

Use recursion: visit left child, then current node, then right child.

4

Run inorder traversal and print the result

Call inorderTraversal with root as argument. Then print the result array values joined by spaces using console.log(result.join(' ')).

DSA Typescript

class TreeNode {
    value: number;
    left: TreeNode | null;
    right: TreeNode | null;

    constructor(value: number) {
        this.value = value;
        this.left = null;
        this.right = null;
    }
}

const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);

const result: number[] = [];

function inorderTraversal(node: TreeNode | null): void {
    if (node !== null) {
        inorderTraversal(node.left);
        result.push(node.value);
        inorderTraversal(node.right);
    }
}

# Your code here

Hint

Call the function with root and print the result array joined by spaces.

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