Tree Traversal Preorder Root Left Right in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Preorder traversal output of a simple binary tree

What is the output of the preorder traversal (Root, Left, Right) of this binary tree?

DSA Typescript

class Node {
  value: number;
  left: Node | null;
  right: Node | null;
  constructor(value: number, left: Node | null = null, right: Node | null = null) {
    this.value = value;
    this.left = left;
    this.right = right;
  }
}

const root = new Node(1, new Node(2), new Node(3));

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

console.log(preorder(root));

A [1, 2, 3]

B [2, 1, 3]

C [3, 2, 1]

D [1, 3, 2]

💡 Hint

Remember preorder means visit root first, then left subtree, then right subtree.

Preorder traversal visits the root node first, then recursively visits the left subtree, and finally the right subtree. Here, root is 1, left child is 2, right child is 3, so output is [1, 2, 3].

❓ Predict Output

intermediate

Preorder traversal output with deeper tree

What is the output of the preorder traversal (Root, Left, Right) of this binary tree?

DSA Typescript

class Node {
  value: number;
  left: Node | null;
  right: Node | null;
  constructor(value: number, left: Node | null = null, right: Node | null = null) {
    this.value = value;
    this.left = left;
    this.right = right;
  }
}

const root = new Node(10, new Node(5, new Node(3), new Node(7)), new Node(15, null, new Node(18)));

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

console.log(preorder(root));

A [5, 3, 7, 10, 15, 18]

B [3, 5, 7, 10, 15, 18]

C [10, 15, 18, 5, 3, 7]

D [10, 5, 3, 7, 15, 18]

💡 Hint

Start from root 10, then go left subtree fully before right subtree.

Preorder visits root first (10), then left subtree (5, 3, 7), then right subtree (15, 18). So output is [10, 5, 3, 7, 15, 18].

🔧 Debug

advanced

Identify the error in preorder traversal implementation

This preorder traversal function is supposed to return an array of node values in preorder. What error will it cause when run?

DSA Typescript

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

A It will cause a stack overflow error due to infinite recursion.

B It will cause a runtime error because result is undefined.

C The output array will be in inorder traversal order, not preorder.

D The output array will be empty because values are never added.

💡 Hint

Check the order in which node values are added to the result array.

The function pushes node values after traversing the left subtree, so it visits left subtree first, then root, then right subtree. This is inorder traversal, not preorder.

🧠 Conceptual

advanced

Number of nodes visited in preorder traversal

Given a binary tree with 7 nodes, how many nodes does the preorder traversal visit?

A 7

B 6

C 8

D Depends on the tree structure

💡 Hint

Preorder traversal visits every node exactly once.

Preorder traversal visits every node in the tree exactly once, so for 7 nodes, it visits 7 nodes.

🚀 Application

expert

Preorder traversal output of an unbalanced tree

What is the preorder traversal output of this unbalanced binary tree?

DSA Typescript

class Node {
  value: number;
  left: Node | null;
  right: Node | null;
  constructor(value: number, left: Node | null = null, right: Node | null = null) {
    this.value = value;
    this.left = left;
    this.right = right;
  }
}

const root = new Node(8, new Node(3, new Node(1), null), new Node(10, null, new Node(14, new Node(13), null)));

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

console.log(preorder(root));

A [1, 3, 8, 10, 14, 13]

B [8, 3, 1, 10, 14, 13]

C [8, 10, 14, 13, 3, 1]

D [8, 3, 10, 1, 14, 13]

💡 Hint

Follow root, then left subtree, then right subtree carefully.

Preorder visits root 8, then left subtree nodes 3 and 1, then right subtree nodes 10, 14, and 13 in order.

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