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.