A binary tree with 7 nodes will have how many null child pointers (empty branches) in its linked representation?

Introduction / Context:
Binary trees are widely used in data structures, and a common way to represent them in memory is through linked nodes. Each node typically has two pointer fields, one for the left child and one for the right child. Many exam questions test the relationship between the number of nodes in a binary tree and the number of null child pointers, because this gives insight into how sparse or full a tree is and helps in reasoning about memory usage and structural properties.

Given Data / Assumptions:

• We have a binary tree with exactly 7 nodes.
• Each node has two child pointers: left and right.
• A null branch means a child pointer that does not point to another node, that is, it is empty or null.
• The tree is a general binary tree, not necessarily a full or complete binary tree, but the null pointer formula still applies.

Concept / Approach:
There is a standard property of binary trees represented with linked nodes: for any non empty binary tree with n nodes, the total number of null child pointers is n + 1. This can be reasoned from the fact that every node (except the root) is pointed to by exactly one child pointer, and there are 2n total child pointer positions. The number of used pointers is n - 1 because the root is not pointed to by any parent. The remaining pointers, which are null, are 2n - (n - 1) = n + 1. This simple relationship holds for every binary tree, regardless of its shape.

Step-by-Step Solution:
Step 1: Note that the binary tree has n = 7 nodes. Step 2: Recognise that each node contributes two child pointer slots, for a total of 2n = 2 * 7 = 14 pointer positions. Step 3: Understand that each node except the root is pointed to exactly once by some parent, so there are n - 1 = 6 non null child pointers. Step 4: Compute the number of null pointers as total child pointers minus non null pointers: 14 - 6 = 8. Step 5: Alternatively apply the direct formula n + 1 = 7 + 1 = 8 for the number of null branches.

Verification / Alternative check:
As a second check, imagine a full binary tree with 7 nodes. A perfect tree of height 2 has one root, two children at level 1, and four children at level 2. The bottom level nodes have no children, so each of the 4 leaf nodes contributes two null child pointers, for a total of 8 null pointers. This visual example matches the n + 1 formula and confirms that the answer of 8 null branches is correct for 7 nodes.

Why Other Options Are Wrong:
Option 6: This would correspond to n - 1 and counts only the non null child pointers, not the null pointers. Option 7: This is equal to n and does not match the derived expression n + 1 for null pointers. Option 5: This has no standard relationship with the number of nodes in the tree and does not arise from the pointer counting argument.

Common Pitfalls:
Students sometimes confuse the number of edges in a tree, which is n - 1, with the number of null pointers, which is n + 1. Another common mistake is to try to reason by listing children manually without a formula, which becomes error prone for larger trees. Remembering the simple relationship that a binary tree with n nodes has n + 1 null child links helps quickly answer many similar questions without drawing the entire structure every time.

Final Answer:
A binary tree with 7 nodes will have 8 null child pointers or empty branches.