Binary Tree Node Structure in DSA Go

Mental Model

A binary tree node holds a value and links to up to two child nodes, left and right.

Analogy: Think of a family tree where each person can have up to two children, one on the left and one on the right.

    [Node]
    /    \
[Left]  [Right]
   ↑       ↑
  left   right

Dry Run Walkthrough

Input: Create a binary tree node with value 10, left child 5, right child 15

Goal: Build a simple binary tree node linking to two children

Step 1: Create root node with value 10

[10] -> left: null, right: null

Why: Start with the main node holding the value

Step 2: Create left child node with value 5 and link to root's left

[10] -> left: [5], right: null

Why: Attach left child to root node

Step 3: Create right child node with value 15 and link to root's right

[10] -> left: [5], right: [15]

Why: Attach right child to root node

Result:

[10] -> left: [5], right: [15]

Annotated Code

DSA Go

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func main() {
    root := &TreeNode{Val: 10}
    root.Left = &TreeNode{Val: 5}
    root.Right = &TreeNode{Val: 15}

    fmt.Printf("%d -> left: %d, right: %d\n", root.Val, root.Left.Val, root.Right.Val)
}

root := &TreeNode{Val: 10}

Create root node with value 10

root.Left = &TreeNode{Val: 5}

Create left child node with value 5 and link to root

root.Right = &TreeNode{Val: 15}

Create right child node with value 15 and link to root

Output

10 -> left: 5, right: 15

Complexity Analysis

Time: O(1) because creating and linking nodes is a constant time operation

Space: O(1) for each node created, space grows linearly with number of nodes

vs Alternative: Compared to arrays, binary tree nodes use pointers to link children, allowing flexible structure without fixed size

Edge Cases

Node with no children

Left and right pointers are null, representing leaf node

DSA Go

// if omitted, left is null
root.Left = &TreeNode{Val: 5}

Node with only one child

One pointer is null, the other points to child node

DSA Go

// can be omitted to have only left child
root.Right = &TreeNode{Val: 15}

When to Use This Pattern

When you need to represent hierarchical data with up to two children per item, use binary tree node structure to link values naturally.

Common Mistakes

Mistake: Forgetting to initialize child nodes before linking
Fix: Always create child nodes with &TreeNode{} before assigning to Left or Right

Summary

It defines a node with a value and two pointers to left and right children.

Use it when building trees where each node can have up to two children.

The key is each node links directly to its children, forming a branching structure.

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