Create a Binary Tree Manually in DSA Go - Execution Trace

Concept Flow - Create a Binary Tree Manually

Create root node

↓

Assign root value

↓

Create left child node

↓

Assign left child value

↓

Create right child node

↓

Assign right child value

↓

Repeat for each child

↓

Tree complete

Start by creating the root node, then create and assign left and right child nodes step-by-step until the tree is built.

Execution Sample

DSA Go

type Node struct {
  value int
  left, right *Node
}

root := &Node{value: 1}
root.left = &Node{value: 2}
root.right = &Node{value: 3}

This code creates a root node with value 1, then manually adds left and right child nodes with values 2 and 3.

Execution Table

Step	Operation	Node Created/Assigned	Pointer Changes	Visual State
1	Create root node	Node(1)	root -> Node(1)	1
2	Create left child node	Node(2)	root.left -> Node(2)	1 / 2
3	Create right child node	Node(3)	root.right -> Node(3)	1 / \ 2 3
4	No more nodes to add	None	No pointer changes	1 / \ 2 3

💡 All nodes created and assigned, tree construction complete

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
root	nil	Node(1)	Node(1)	Node(1)	Node(1)
root.left	nil	nil	Node(2)	Node(2)	Node(2)
root.right	nil	nil	nil	Node(3)	Node(3)

Key Moments - 3 Insights

Why do we assign root.left and root.right separately instead of during root creation?

Because in manual tree creation, child nodes are created after the root exists. Step 2 and 3 in the execution_table show root is created first, then children are assigned.

What does the visual state mean after each step?

It shows the tree structure with node values and their left/right children. For example, after step 3, the tree has root 1 with left child 2 and right child 3.

Why is root initially nil and then assigned a node?

At start, root points to nothing (nil). Step 1 creates the root node and assigns it to root, as shown in variable_tracker.

Visual Quiz - 3 Questions

Look at the execution_table at step 2, what pointer changes occur?

A root.left points to Node(2)

B root.right points to Node(2)

C root points to Node(2)

D No pointer changes

💡 Hint

Check the 'Pointer Changes' column in execution_table row for step 2

At which step does root.right get assigned a node?

A Step 1

B Step 3

C Step 2

D Step 4

💡 Hint

Look at the 'Operation' and 'Pointer Changes' columns in execution_table

If we add a left child to Node(2), how would the visual state change after step 4?

A 1 with no children

B 1 with right child 3 only

C 1 with left child 2, which has left child added

D Only Node(2) changes, root stays nil

💡 Hint

Visual state shows tree structure; adding a child to Node(2) extends the left subtree

Concept Snapshot

Create a binary tree manually by:
- Define Node struct with value, left, right pointers
- Create root node first
- Assign left and right children separately
- Repeat for deeper levels
- Visualize tree as nodes connected by pointers

Full Transcript

To create a binary tree manually, start by defining a node structure with a value and pointers to left and right children. First, create the root node and assign it to a variable. Then create left and right child nodes and assign them to the root's left and right pointers respectively. Repeat this process for each child node to build the tree. The execution table shows each step with the node created, pointer changes, and the visual tree state. The variable tracker shows how root and its children pointers change after each step. Key moments clarify why children are assigned after root creation and how the visual state represents the tree structure. The visual quiz tests understanding of pointer assignments and tree structure changes.

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