Binary Tree Node Structure in DSA Go - Execution Trace

Concept Flow - Binary Tree Node Structure

Create new node

↓

Assign data value

↓

Set left child pointer to nil

↓

Set right child pointer to nil

↓

Node ready for linking in tree

This flow shows how a binary tree node is created with data and two child pointers initialized to nil.

Execution Sample

DSA Go

type Node struct {
    data  int
    left  *Node
    right *Node
}

root := &Node{data: 10, left: nil, right: nil}

This code defines a binary tree node structure and creates a root node with data 10 and no children.

Execution Table

Step	Operation	Node Created	Left Pointer	Right Pointer	Visual State
1	Create new node	Node(data=10)	undefined	undefined	┌────────┐ │ data:10│ │ left:∅ │ │right:∅ │ └────────┘
2	Assign data value	Node(data=10)	undefined	undefined	┌────────┐ │ data:10│ │ left:∅ │ │right:∅ │ └────────┘
3	Set left pointer to nil	Node(data=10)	nil	undefined	┌────────┐ │ data:10│ │ left:∅ │ │right:∅ │ └────────┘
4	Set right pointer to nil	Node(data=10)	nil	nil	┌────────┐ │ data:10│ │ left:∅ │ │right:∅ │ └────────┘
5	Node ready for linking	Node(data=10)	nil	nil	┌────────┐ │ data:10│ │ left:∅ │ │right:∅ │ └────────┘

💡 All pointers initialized to nil, node is ready to be linked in the tree.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
node.data	undefined	10	10	10	10	10
node.left	undefined	undefined	undefined	nil	nil	nil
node.right	undefined	undefined	undefined	undefined	nil	nil

Key Moments - 2 Insights

Why are left and right pointers set to nil initially?

Because in the execution_table rows 3 and 4, we see left and right pointers assigned nil to indicate no children yet, preventing accidental references.

Is the data value assigned before or after pointers are set?

Data is assigned first as shown in step 2, then pointers are set to nil in steps 3 and 4, ensuring the node has valid data before linking.

Visual Quiz - 3 Questions

Look at the execution_table at step 1, what is the state of the left pointer?

A nil

B points to another node

C undefined

D points to itself

💡 Hint

Check the 'Left Pointer' column at step 1 in execution_table.

At which step do both left and right pointers become nil?

A Step 2

B Step 4

C Step 3

D Step 5

💡 Hint

Look at the 'Left Pointer' and 'Right Pointer' columns in execution_table rows.

If we skip setting right pointer to nil, what would be the final state of node.right?

A undefined or garbage value

B nil

C points to left child

D points to data

💡 Hint

Refer to variable_tracker for node.right initialization and step 4 in execution_table.

Concept Snapshot

Binary Tree Node Structure:
- Each node holds data and two pointers: left and right.
- Pointers start as nil (no children).
- Nodes link by assigning these pointers.
- This structure forms the basis for binary trees.

Full Transcript

This visual trace shows how a binary tree node is created in Go. First, a new node is made with data 10. Then, its left and right pointers are set to nil, meaning no children yet. The node is now ready to be linked into a binary tree. The execution table tracks each step and the node's state. The variable tracker shows data and pointers before and after each step. Key moments clarify why pointers start as nil and the order of assignments. The quiz tests understanding of pointer initialization and node state.

Practice

1. What is the correct way to define a node structure for a binary tree in Go?

easy

A. type Node struct { Value int; Left Node; Right Node }

B. type Node struct { Value int; Left *Node; Right *Node }

C. type Node struct { Value int; Left int; Right int }

D. type Node struct { Value *int; Left *int; Right *int }

Solution

1. Step 1: Understand binary tree node components

   A binary tree node stores a value and pointers to left and right child nodes.

2. Step 2: Check pointer usage in Go struct

   Using pointers (*Node) for Left and Right allows linking nodes without copying entire structs.

3. Final Answer:

   type Node struct { Value int; Left *Node; Right *Node } -> Option B

4. Quick Check:

   Node with pointers = type Node struct { Value int; Left *Node; Right *Node } [OK]

Hint: Use pointers for child nodes to avoid copying structs [OK]

Common Mistakes:

• Using non-pointer types for child nodes causes infinite recursion
• Using int instead of *Node for children
• Using pointers to int instead of pointers to Node

2. Which of the following is the correct syntax to create a new binary tree node with value 10 in Go?

easy

A. node := Node{Value: 10, Left: nil, Right: nil}

B. node := Node(10, nil, nil)

C. node := &Node{Value: 10, Left: nil, Right: nil}

D. node := new(Node{10, nil, nil})

Solution

1. Step 1: Understand node creation with pointers

   To create a pointer to a Node struct, use &Node{...} syntax.

2. Step 2: Check each option's syntax

   node := &Node{Value: 10, Left: nil, Right: nil} correctly creates a pointer to a Node with value 10 and nil children.

3. Final Answer:

   node := &Node{Value: 10, Left: nil, Right: nil} -> Option C

4. Quick Check:

   Pointer node creation uses &Node{...} [OK]

Hint: Use &Node{...} to create pointer nodes [OK]

Common Mistakes:

• Forgetting & to create pointer
• Using new() incorrectly with struct literals
• Calling Node() like a function

3. What will be the output of the following Go code snippet?

package main
import "fmt"

type Node struct {
  Value int
  Left  *Node
  Right *Node
}

func main() {
  root := &Node{Value: 1}
  root.Left = &Node{Value: 2}
  root.Right = &Node{Value: 3}
  fmt.Println(root.Left.Value)
}

medium

A. 2

B. 1

C. 3

D. Compilation error

Solution

1. Step 1: Analyze node assignments

   root.Left is assigned a node with Value 2, root.Right with Value 3, root.Value is 1.

2. Step 2: Check printed value

   fmt.Println(root.Left.Value) prints the Value of the left child node, which is 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   root.Left.Value = 2 [OK]

Hint: Print child node's Value to get assigned integer [OK]

Common Mistakes:

• Printing root.Value instead of root.Left.Value
• Confusing left and right child values
• Expecting compilation error due to missing import

4. Identify the error in the following Go code defining a binary tree node:

type Node struct {
  Value int
  Left Node
  Right *Node
}

func main() {
  root := &Node{Value: 5}
  root.Left = Node{Value: 3}
}

medium

A. Left child should be a pointer to Node, not Node value

B. Right child should not be a pointer

C. Value field must be a pointer to int

D. Cannot assign struct literal to Left field

Solution

1. Step 1: Check struct field types

   Left is declared as Node (value), Right as *Node (pointer). This causes recursive struct size issues.

2. Step 2: Understand pointer necessity

   Binary tree nodes must use pointers for children to avoid infinite size and allow nil values.

3. Final Answer:

   Left child should be a pointer to Node, not Node value -> Option A

4. Quick Check:

   Child nodes must be pointers [OK]

Hint: Use *Node for all child fields to avoid infinite struct size [OK]

Common Mistakes:

• Mixing pointer and value types for children
• Expecting Go to allow recursive value fields
• Ignoring nil possibility for children

5. Given the binary tree node structure below, how would you write a function in Go to insert a new value into a binary search tree?

type Node struct {
  Value int
  Left  *Node
  Right *Node
}

Choose the correct approach.

hard

A. Insert new value at root and push existing nodes down

B. Always insert new value as left child without comparison

C. Use a loop to find nil child and insert without comparison

D. Recursively compare new value with current node; insert left if smaller, right if larger; create new node if nil child

Solution

1. Step 1: Understand binary search tree insertion logic

   New values less than current node go to left subtree; greater go to right subtree.

2. Step 2: Use recursion to find correct insertion point

   Recursively traverse left or right child until a nil child is found, then insert new node there.

3. Final Answer:

   Recursively compare new value with current node; insert left if smaller, right if larger; create new node if nil child -> Option D

4. Quick Check:

   BST insertion uses recursive comparison [OK]

Hint: Compare and recurse left/right until nil, then insert [OK]

Common Mistakes:

• Inserting without comparison breaks BST property
• Always inserting on one side causes imbalance
• Not handling nil child case properly