In the absolute worst case, a binary tree with N elements would be like a linked list.
Hence, there would be N levels, and a search would take N traversals.
~ Root ~
____
| 42 |
|____|
/ \
____/ \
| 13 | X
|____|
/ \
____/ \
| 11 | X
|____|
/ \
/ \
... X
That’s why it’s O(N) in the worst case.
And this is why we need to balance the trees to achieve O(log N) search.