# Exercise 12.3.2

Suppose that we construct a binary search tree by repeatedly inserting distinct values into the tree. Argue that the number of nodes examined in searching for a value in the tree is one plus the number of nodes examined when the value was first inserted into the tree.

With the current implementation of TREE-INSERT (one that does not self-balance), we are looking for a path from to root to the parent of the node we are about to insert. Let's assume that that number is $n$. When we subsequently search, we are going to walk exactly the same path (there is no other option, really) until we reach the same parent. Then we will examine one more node, which is the node we inserted previously, that is $n+1$ in total.

This argument holds, since we're always inserting distinct values, and there is only one possible path to resulting to the value we're searching for.