Binary Tree Node Structure in DSA C - Execution Trace

Concept Flow - Binary Tree Node Structure

Create new node

↓

Assign data value

↓

Set left child pointer to null

↓

Set right child pointer to null

↓

Node ready for use in tree

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

Execution Sample

DSA C

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

Defines a binary tree node with an integer data and two pointers for left and right children.

Execution Table

Step	Operation	Node Field	Value Assigned	Visual State
1	Create node variable	Node	Allocated memory	∅ (empty node)
2	Assign data	data	10	Node(data:10, left:undefined, right:undefined)
3	Initialize left pointer	left	∅ (null)	Node(data:10, left:∅, right:undefined)
4	Initialize right pointer	right	∅ (null)	Node(data:10, left:∅, right:∅)
5	Node ready	Node	data=10, left=∅, right=∅	┌────────┐ │ data:10│ │ left:∅ │ │ right:∅│ └────────┘

💡 All fields assigned, node structure complete and ready for tree insertion.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
Node.data	undefined	10	10	10	10
Node.left	undefined	undefined	∅	∅	∅
Node.right	undefined	undefined	undefined	∅	∅

Key Moments - 2 Insights

Why do left and right pointers start as null?

Because in the execution_table at steps 3 and 4, both pointers are set to ∅ (null) to show no children yet.

What does the visual state show after assigning data?

At step 2, the visual state shows the node with data 10 but left and right pointers still undefined, meaning no children connected.

Visual Quiz - 3 Questions

Look at the execution_table at step 2, what is the value of Node.data?

A 10

B null

C undefined

D 0

💡 Hint

Check the 'Value Assigned' column at step 2 in execution_table.

At which step are both left and right pointers set to null?

A Step 3

B Step 4

C Step 2

D Step 5

💡 Hint

Look at the 'Operation' and 'Node Field' columns in execution_table for pointer initialization.

If we skip setting left pointer to null, what would variable_tracker show after step 3?

A Node.left = 10

B Node.left = ∅

C Node.left = undefined

D Node.left = Node.right

💡 Hint

Refer to variable_tracker row for Node.left after step 3.

Concept Snapshot

Binary Tree Node Structure:
- Contains data, left pointer, right pointer
- Left and right pointers start as null
- Used to build binary trees
- Each node links to up to two children
- Basic building block for binary tree operations

Full Transcript

A binary tree node has three parts: data, left pointer, and right pointer. We create a node, assign data, and set both pointers to null to show no children. This node can then be linked into a binary tree structure.

Practice

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

easy

A. Two data fields and no pointers

B. Only a data field without any pointers

C. A data field and a pointer to the parent node only

D. A data field and two pointers to left and right child nodes

Solution

1. Step 1: Understand binary tree node components

   A binary tree node must hold data and links to its children to form the tree structure.

2. Step 2: Identify correct structure

   The node typically has one data field and two pointers: one for the left child and one for the right child.

3. Final Answer:

   A data field and two pointers to left and right child nodes -> Option D

4. Quick Check:

   Binary tree node = data + left + right pointers [OK]

Hint: Remember: binary tree nodes have two child pointers [OK]

Common Mistakes:

• Thinking nodes have no pointers
• Confusing parent pointer with child pointers
• Assuming multiple data fields

2. Which of the following is the correct C syntax to define a binary tree node structure?

easy

A. struct Node { int data; struct Node left; struct Node right; };

B. struct Node { int data; Node* left; Node* right; };

C. struct Node { int data; struct Node* left; struct Node* right; };

D. struct Node { int data; struct Node* parent; };

Solution

1. Step 1: Check pointer usage in struct

   In C, pointers to the same struct type must be declared with 'struct Node*' inside the struct.

2. Step 2: Identify correct syntax

   struct Node { int data; struct Node* left; struct Node* right; }; correctly uses 'struct Node*' for left and right child pointers.

3. Final Answer:

   struct Node { int data; struct Node* left; struct Node* right; }; -> Option C

4. Quick Check:

   Use 'struct Node*' for pointers inside struct [OK]

Hint: Use 'struct Node*' for pointers inside struct in C [OK]

Common Mistakes:

• Omitting 'struct' keyword before Node*
• Using struct Node instead of pointer causing infinite size
• Confusing parent pointer with child pointers

3. Given the following code, what will be printed?

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

int main() {
  struct Node n1 = {10, NULL, NULL};
  struct Node n2 = {20, NULL, NULL};
  struct Node root = {5, &n1, &n2};
  printf("%d %d %d", root.data, root.left->data, root.right->data);
  return 0;
}

medium

A. 5 10 20

B. 10 5 20

C. 5 0 0

D. Compilation error

Solution

1. Step 1: Understand node initialization

   n1 has data 10, n2 has data 20, root has data 5 with left pointing to n1 and right pointing to n2.

2. Step 2: Trace printf output

   root.data is 5, root.left->data is 10, root.right->data is 20, so output is '5 10 20'.

3. Final Answer:

   5 10 20 -> Option A

4. Quick Check:

   Print root and children data = 5 10 20 [OK]

Hint: Follow pointers to child nodes for data [OK]

Common Mistakes:

• Confusing data values of nodes
• Assuming NULL pointers print zero
• Mixing order of print arguments

4. Identify the error in this binary tree node definition:

struct Node {
  int data;
  struct Node left;
  struct Node right;
};

medium

A. Left and right should be pointers, not structs

B. Missing semicolon after struct definition

C. Data field should be a pointer

D. Struct name should be lowercase

Solution

1. Step 1: Analyze struct member types

   Left and right are declared as struct Node, which causes infinite size because each contains two more structs recursively.

2. Step 2: Correct pointer usage

   They should be pointers (struct Node*) to avoid infinite recursion and allow linking nodes.

3. Final Answer:

   Left and right should be pointers, not structs -> Option A

4. Quick Check:

   Child nodes must be pointers to avoid infinite size [OK]

Hint: Use pointers for child nodes to prevent infinite size [OK]

Common Mistakes:

• Declaring child nodes as structs instead of pointers
• Ignoring compiler errors about struct size
• Thinking data field needs to be pointer

5. You want to create a binary tree node in C that stores a character and has pointers to left and right children. Which of the following struct definitions correctly achieves this and allows dynamic node creation?

hard

A. struct Node { char data; struct Node left; struct Node right; }; struct Node* createNode(char c) { struct Node newNode; newNode.data = c; newNode.left = NULL; newNode.right = NULL; return &newNode; }

B. struct Node { char data; struct Node* left; struct Node* right; }; struct Node* createNode(char c) { struct Node* newNode = malloc(sizeof(struct Node)); newNode->data = c; newNode->left = NULL; newNode->right = NULL; return newNode; }

C. struct Node { char data; struct Node* left; struct Node* right; }; struct Node createNode(char c) { struct Node newNode; newNode.data = c; newNode.left = NULL; newNode.right = NULL; return newNode; }

D. struct Node { char data; struct Node* left; struct Node* right; }; struct Node* createNode(char c) { struct Node newNode; newNode.data = c; newNode.left = NULL; newNode.right = NULL; return newNode; }

Solution

1. Step 1: Check struct definition

   struct Node { char data; struct Node* left; struct Node* right; }; struct Node* createNode(char c) { struct Node* newNode = malloc(sizeof(struct Node)); newNode->data = c; newNode->left = NULL; newNode->right = NULL; return newNode; } correctly defines pointers for left and right children and uses char data type.

2. Step 2: Verify dynamic node creation

   struct Node { char data; struct Node* left; struct Node* right; }; struct Node* createNode(char c) { struct Node* newNode = malloc(sizeof(struct Node)); newNode->data = c; newNode->left = NULL; newNode->right = NULL; return newNode; } uses malloc to allocate memory dynamically and initializes pointers to NULL, returning a pointer to the new node.

3. Final Answer:

   Correct struct with pointers and dynamic allocation using malloc -> Option B

4. Quick Check:

   Dynamic node creation requires malloc and pointer returns [OK]

Hint: Use malloc and pointers for dynamic binary tree nodes [OK]

Common Mistakes:

• Returning address of local variable (invalid after function ends)
• Using struct members instead of pointers for children
• Not initializing child pointers to NULL