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

Tree Traversal Preorder Root Left Right

📖 Scenario: You are working with a simple family tree where each person can have up to two children. You want to visit each person starting from the oldest ancestor, then their left child, then their right child.

🎯 Goal: Build a program that creates a tree structure and prints the names of people in preorder traversal order: root first, then left child, then right child.

📋 What You'll Learn

Create a tree node class with name, left, and right properties

Create a tree with exactly three nodes: "Grandparent", "Parent", and "Child"

Write a preorder traversal function called preorderTraversal that visits nodes in root-left-right order

Print the names of the nodes in preorder traversal separated by spaces

💡 Why This Matters

🌍 Real World

Tree traversal is used in many real-world applications like file system navigation, organization charts, and parsing expressions.

💼 Career

Understanding tree traversal is essential for software developers working with hierarchical data, databases, and algorithms.

1

Create TreeNode class and build the tree

Create a class called TreeNode with a constructor that takes a name string and sets left and right to null. Then create three nodes: grandparent with name "Grandparent", parent with name "Parent", and child with name "Child". Connect grandparent.left to parent and parent.left to child.

DSA Typescript

// Your code here

Hint

Define the TreeNode class with name, left, and right. Then create the three nodes and link them as described.

2

Create a helper array to store traversal result

Create an empty array called result of type string[] to store the names of nodes during traversal.

DSA Typescript

class TreeNode {
    name: string;
    left: TreeNode | null;
    right: TreeNode | null;

    constructor(name: string) {
        this.name = name;
        this.left = null;
        this.right = null;
    }
}

const grandparent = new TreeNode("Grandparent");
const parent = new TreeNode("Parent");
const child = new TreeNode("Child");

grandparent.left = parent;
parent.left = child;

// Create an empty array called result to store traversal names
// Your code here

Hint

Just create an empty array named result to hold the names as strings.

3

Write preorder traversal function

Write a function called preorderTraversal that takes a node of type TreeNode | null. If node is not null, push node.name to result, then recursively call preorderTraversal on node.left and node.right.

DSA Typescript

class TreeNode {
    name: string;
    left: TreeNode | null;
    right: TreeNode | null;

    constructor(name: string) {
        this.name = name;
        this.left = null;
        this.right = null;
    }
}

const grandparent = new TreeNode("Grandparent");
const parent = new TreeNode("Parent");
const child = new TreeNode("Child");

grandparent.left = parent;
parent.left = child;

const result: string[] = [];

// Write preorderTraversal function here
// Your code here

Hint

Use a function that checks if the node is not null, then adds the name to result, then calls itself on left and right children.

4

Run traversal and print result

Call preorderTraversal with grandparent as argument. Then print the result array joined by spaces using console.log(result.join(" ")).

DSA Typescript

class TreeNode {
    name: string;
    left: TreeNode | null;
    right: TreeNode | null;

    constructor(name: string) {
        this.name = name;
        this.left = null;
        this.right = null;
    }
}

const grandparent = new TreeNode("Grandparent");
const parent = new TreeNode("Parent");
const child = new TreeNode("Child");

grandparent.left = parent;
parent.left = child;

const result: string[] = [];

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

// Call preorderTraversal and print the result
// Your code here

Hint

Call the traversal starting from grandparent and print the names joined by spaces.

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