Binary Tree Node Structure in DSA Python - Execution Trace

Concept Flow - Binary Tree Node Structure

Create new node

↓

Set node data

↓

Set left child pointer to None

↓

Set right child pointer to None

↓

Node ready to link in tree

This flow shows how a binary tree node is created with data and two pointers set to None, ready to be linked.

Execution Sample

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

This code creates a binary tree node with data and two child pointers initialized to None.

Execution Table

Step	Operation	Node Data	Left Pointer	Right Pointer	Visual State
1	Create node object	Uninitialized	Uninitialized	Uninitialized	No node created yet
2	Assign data = 10	10	Uninitialized	Uninitialized	Node data set to 10
3	Set left = None	10	None	Uninitialized	Left pointer set to None
4	Set right = None	10	None	None	Right pointer set to None
5	Node ready	10	None	None	┌────────┐ │ data:10│ │ left:∅ │ │right:∅ │ └────────┘

💡 Node fully initialized with data and both child pointers set to None

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
node.data	Uninitialized	10	10	10	10
node.left	Uninitialized	Uninitialized	None	None	None
node.right	Uninitialized	Uninitialized	Uninitialized	None	None

Key Moments - 2 Insights

Why are left and right pointers set to None initially?

Because the node starts without children, setting left and right to None shows no child nodes exist yet, as seen in execution_table steps 3 and 4.

What does the visual state represent at the end?

It shows the node as a box with data and pointers, where ∅ means None (no child), matching the final row in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the value of node.left?

A 10

B None

C Uninitialized

D Node object

💡 Hint

Check the 'Left Pointer' column at step 3 in execution_table

At which step are both left and right pointers set to None?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

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

If we skip setting right pointer to None, what would be the final right pointer value?

A Uninitialized

B None

C Data value

D Node object

💡 Hint

Refer to variable_tracker for node.right values after each step

Concept Snapshot

Binary Tree Node Structure:
- Create node with data
- Initialize left and right pointers to None
- Pointers show child nodes (None means no child)
- Node ready to link in tree
- Visual: box with data and two pointers

Full Transcript

A binary tree node is created by making an object with a data value and two pointers called left and right. Both pointers start as None, meaning the node has no children yet. This node can then be linked into a larger tree structure. The code example shows a class Node with an __init__ method that sets these values. Step-by-step, the node data is assigned, then left and right pointers are set to None. The final visual shows a box with data and pointers marked as empty (∅).

Practice

1. What does a typical binary tree node contain in its structure?

easy

A. A value and a reference to the parent node only

B. Only a value and a reference to the next node

C. A value, a reference to the left child, and a reference to the right child

D. A value and a list of all children nodes

Solution

1. Step 1: Understand binary tree node components

   A binary tree node typically stores a value and has two child references: left and right.

2. Step 2: Compare options with definition

   A value, a reference to the left child, and a reference to the right child matches this structure exactly, while others describe different data structures.

3. Final Answer:

   A value, a reference to the left child, and a reference to the right child -> Option C

4. Quick Check:

   Binary tree node = value + left + right [OK]

Hint: Binary tree nodes always have left and right child pointers [OK]

Common Mistakes:

• Confusing binary tree node with linked list node
• Thinking nodes have parent pointers by default
• Assuming children are stored in a list

2. Which of the following is the correct Python syntax to define a binary tree node class with attributes val, left, and right?

easy

A. class Node:
def __init__(self, val):
val = self
left = None
right = None

B. class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None

C. class Node:
def __init__(self):
self.val = None
self.left = None
self.right = None

D. class Node:
def __init__(self, val, left, right):
self.val = val
self.left = left
self.right = right

Solution

1. Step 1: Check constructor parameters and attribute assignments

   class Node: def __init__(self, val): self.val = val self.left = None self.right = None correctly initializes val and sets left and right to None by default.

2. Step 2: Validate syntax correctness

   class Node: def __init__(self, val): val = self left = None right = None incorrectly assigns variables; class Node: def __init__(self): self.val = None self.left = None self.right = None misses val parameter; class Node: def __init__(self, val, left, right): self.val = val self.left = left self.right = right requires left and right on init which is less common.

3. Final Answer:

   class Node: def __init__(self, val): self.val = val self.left = None self.right = None -> Option B

4. Quick Check:

   Constructor sets val and initializes children to None [OK]

Hint: Init val, set left and right to None inside __init__ [OK]

Common Mistakes:

• Not using self to assign attributes
• Forgetting to initialize left and right
• Incorrect parameter order or missing parameters

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

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

root = Node(10)
root.left = Node(5)
root.right = Node(15)
print(root.left.val, root.right.val)

medium

A. 15 5

B. 10 5

C. None None

D. 5 15

Solution

1. Step 1: Trace node creation and assignments

   Root node has value 10; left child is assigned a node with value 5; right child assigned node with value 15.

2. Step 2: Print left and right child values

   Printing root.left.val and root.right.val outputs 5 and 15 respectively.

3. Final Answer:

   5 15 -> Option D

4. Quick Check:

   Left child val = 5, Right child val = 15 [OK]

Hint: Print left.val and right.val after assignment to get child values [OK]

Common Mistakes:

• Printing root.val instead of children
• Confusing left and right child values
• Assuming children are None without assignment

4. Identify the error in the following binary tree node class definition:

class Node:
    def __init__(self, val):
        val = self.val
        left = None
        right = None

medium

A. Attributes are assigned incorrectly; should use self.val, self.left, self.right

B. Missing return statement in __init__ method

C. Class name should be lowercase

D. No error; code is correct

Solution

1. Step 1: Review attribute assignments in __init__

   Attributes must be assigned to self to belong to the instance, e.g., self.val = val.

2. Step 2: Identify incorrect assignments

   Code assigns val = self.val which reverses assignment and does not use self for left and right.

3. Final Answer:

   Attributes must be assigned using self, not local variables -> Option A

4. Quick Check:

   Use self.attr = value to set instance variables [OK]

Hint: Always assign attributes with self.attribute = value [OK]

Common Mistakes:

• Assigning variables without self prefix
• Reversing assignment order
• Expecting return in __init__

5. Given the following binary tree node class, how would you create a tree with root value 1, left child 2, and right child 3, then print the root and its children values?

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

hard

A. root = Node(1)
root.left = Node(2)
root.right = Node(3)
print(root.val, root.left.val, root.right.val)

B. root = Node(1, Node(2), Node(3))
print(root.val, root.left.val, root.right.val)

C. root = Node()
root.val = 1
root.left = 2
root.right = 3
print(root.val, root.left, root.right)

D. root = Node(1)
root.left = 2
root.right = 3
print(root.val, root.left, root.right)

Solution

1. Step 1: Create root node with value 1

   Use Node(1) to create root with val=1 and children None.

2. Step 2: Assign left and right children as Node instances

   Assign root.left = Node(2) and root.right = Node(3) to create child nodes properly.

3. Step 3: Print values of root and children

   Print root.val, root.left.val, root.right.val to get 1 2 3.

4. Final Answer:

   Use separate Node instances for children and print their val attributes -> Option A

5. Quick Check:

   Children must be Node objects, not integers [OK]

Hint: Assign children as Node objects, then print their val attributes [OK]

Common Mistakes:

• Assigning children as integers instead of Node objects
• Using constructor with extra parameters (not defined)
• Printing child nodes directly without .val