Create a Binary Tree Manually in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a struct for a binary tree node.

DSA Go

type Node struct {
    value int
    left  *[1]
    right *Node
}

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

ANode

BTree

CNodePtr

DBinaryNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different struct name for left or right child.

Forgetting the pointer (*) before the struct name.

The left child pointer should be of type *Node to refer to the same struct type.

2fill in blank

medium

Complete the code to create a new node with value 10.

DSA Go

root := &Node{value: [1], left: nil, right: nil}

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

A0

B10

Cnil

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting value to 0 or nil instead of 10.

Confusing the pointer fields with the value field.

The root node should have the value 10 as specified.

3fill in blank

hard

Fix the error in assigning the left child node with value 5.

DSA Go

root.left = &Node{value: [1], left: nil, right: nil}

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

A0

B10

Cnil

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using the root value 10 instead of 5 for the left child.

Assigning nil or zero instead of the correct value.

The left child node should have the value 5 as specified.

4fill in blank

hard

Fill both blanks to create the right child node with value 15 and no children.

DSA Go

root.right = &Node{value: [1], left: [2], right: nil}

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

A15

Bnil

C0

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or root instead of 15 for value.

Using 0 or root instead of nil for left child pointer.

The right child node has value 15 and no left child (nil).

5fill in blank

hard

Fill all three blanks to create a leaf node with value 7 and no children.

DSA Go

leaf := &Node{value: [1], left: [2], right: [3]

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

A7

Bnil

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of nil for child pointers.

Setting wrong value for the leaf node.

A leaf node has a value and both left and right pointers are nil.

Practice

1. What is the first step to manually create a binary tree in Go?

easy

A. Define a struct type for the tree node with data and pointers

B. Write a function to traverse the tree

C. Initialize an empty slice

D. Import the fmt package

Solution

1. Step 1: Define the node structure

   To create a binary tree manually, you first need a struct that represents each node with fields for data and pointers to left and right children.

2. Step 2: Use this struct to create nodes and link them

   After defining the struct, you can create nodes and connect them to form the tree.

3. Final Answer:

   Define a struct type for the tree node with data and pointers -> Option A

4. Quick Check:

   Node struct definition = A [OK]

Hint: Start by defining the node struct with left and right pointers [OK]

Common Mistakes:

• Trying to create nodes without defining the struct first
• Confusing slices with tree nodes
• Skipping pointer fields in the struct

2. Which of the following is the correct syntax to declare a binary tree node struct in Go?

easy

A. class Node { int data; Node left; Node right }

B. struct Node { int data; Node* left; Node* right }

C. Node := struct { data int; left Node; right Node }

D. type Node struct { data int; left *Node; right *Node }

Solution

1. Step 1: Understand Go struct syntax

   In Go, structs are declared with the keyword 'type' followed by the struct name and fields inside curly braces. Pointers use '*'.

2. Step 2: Match the correct syntax

   type Node struct { data int; left *Node; right *Node } correctly uses 'type Node struct { data int; left *Node; right *Node }' which is valid Go syntax for a binary tree node.

3. Final Answer:

   type Node struct { data int; left *Node; right *Node } -> Option D

4. Quick Check:

   Go struct with pointers = C [OK]

Hint: Remember Go uses 'type' and '*' for pointers in structs [OK]

Common Mistakes:

• Using C/C++ or Java syntax instead of Go
• Missing the '*' for pointer fields
• Using := inside type declaration

3. What will be the output of this Go code that manually creates a binary tree and prints the root node's data?

package main
import "fmt"
type Node struct {
  data int
  left *Node
  right *Node
}
func main() {
  root := &Node{data: 10}
  root.left = &Node{data: 5}
  root.right = &Node{data: 15}
  fmt.Println(root.data)
}

medium

A. 10

B. 5

C. 15

D. Compilation error

Solution

1. Step 1: Understand node creation

   The root node is created with data 10, and its left and right children are assigned 5 and 15 respectively.

2. Step 2: Identify printed value

   The code prints root.data, which is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   root.data = 10 [OK]

Hint: Print root.data to get the root node's value [OK]

Common Mistakes:

• Printing left or right data instead of root
• Confusing pointer dereferencing syntax
• Expecting traversal output instead of root data

4. Identify the error in this Go code that tries to create a binary tree node:

type Node struct {
  data int
  left *Node
  right *Node
}
func main() {
  root := Node{data: 10}
  root.left = Node{data: 5}
  root.right = Node{data: 15}
}

medium

A. Missing semicolon after struct definition

B. Data field should be string, not int

C. Cannot assign struct to pointer field without & operator

D. No error, code is correct

Solution

1. Step 1: Check pointer assignments

   The left and right fields are pointers (*Node), but the code assigns Node structs directly without using & to get their address.

2. Step 2: Correct pointer assignment

   To fix, use &Node{data: 5} and &Node{data: 15} to assign pointers.

3. Final Answer:

   Cannot assign struct to pointer field without & operator -> Option C

4. Quick Check:

   Pointer fields need & before struct [OK]

Hint: Use & to assign address to pointer fields [OK]

Common Mistakes:

• Assigning struct value directly to pointer field
• Ignoring pointer type in struct fields
• Assuming Go auto-converts struct to pointer

5. Given the following Go code to create a binary tree manually, what is the correct way to add a new node with value 7 as the left child of the node with value 5?

type Node struct {
  data int
  left *Node
  right *Node
}
func main() {
  root := &Node{data: 10}
  root.left = &Node{data: 5}
  root.right = &Node{data: 15}
  // Add new node here
}

hard

A. root.right.right = &Node{data: 7}

B. root.left.left = &Node{data: 7}

C. root.right.left = &Node{data: 7}

D. root.left.right = &Node{data: 7}

Solution

1. Step 1: Locate node with value 5

   The node with value 5 is root.left.

2. Step 2: Add new node as left child of node 5

   To add a left child, assign root.left.left = &Node{data: 7}.

3. Final Answer:

   root.left.left = &Node{data: 7} -> Option B

4. Quick Check:

   Left child of node 5 = D [OK]

Hint: Use root.left.left to add left child to node 5 [OK]

Common Mistakes:

• Adding new node to wrong child pointer
• Confusing left and right pointers
• Assigning to root.right instead of root.left