Binary Tree Node Structure in DSA Python

Mental Model

A binary tree node holds a value and links to two child nodes, left and right, which can be empty.

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]
   null   null

Dry Run Walkthrough

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

Goal: Build a simple binary tree node with two children linked correctly

Step 1: Create root node with value 10

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

Why: Start with the main node that holds 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 to build the left branch

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 to build the right branch

Result:

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

Annotated Code

DSA Python

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

# Driver code to create the tree node as per dry run
root = TreeNode(10)
root.left = TreeNode(5)
root.right = TreeNode(15)

# Function to print tree node and its children

def print_tree_node(node):
    left_val = node.left.val if node.left else 'null'
    right_val = node.right.val if node.right else 'null'
    print(f"[{node.val}] -> left: [{left_val}], right: [{right_val}]")

print_tree_node(root)

root = TreeNode(10)

Create root node with value 10

root.left = TreeNode(5)

Create left child node with value 5 and link to root

root.right = TreeNode(15)

Create right child node with value 15 and link to root

print_tree_node(root)

Print root node and its children values

Output

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

Complexity Analysis

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

Space: O(1) because only a fixed number of nodes are created

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

Edge Cases

Node with no children

Left and right pointers are null, representing leaf node

DSA Python

self.left = left

Node with only left child

Right pointer is null, left points to child node

DSA Python

self.right = right

Node with only right child

Left pointer is null, right points to child node

DSA Python

self.right = right

When to Use This Pattern

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

Common Mistakes

Mistake: Forgetting to initialize left and right pointers to null
Fix: Always set left and right to None in the constructor if no child is given

Mistake: Confusing left and right child assignments
Fix: Assign left child to left pointer and right child to right pointer explicitly

Summary

It creates a node that holds a value and links to two child nodes in a binary tree.

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

The key is each node stores its own value and pointers to left and right children, allowing tree structure.

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