Solution:

Answer: $14$

Let the people be $A, B, C, D, E$ so that their heights are in that order, with $A$ the tallest and $E$ the shortest. We will do casework based on the position of $C$.

- Case 1: $C$ is in the middle. Then, $B$ must be on one of the two ends, for two choices. This leaves only one choice for $D$—the other end. Then, we know the positions of $A$ and $E$ since $A$ cannot neighbor $B$ and $E$ cannot neighbor $D$. So we have $2$ options for this case.

- Case 2: $C$ is in the second or fourth spot. Then, we have two choices for the position of $C$. Without loss of generality, let $C$ be in the second spot. Then, the first and third spots must be $A$ and $E$, giving us two options. This fixes the positions of $B$ and $D$, so we have a total of $2 \times 2 = 4$ options for this case.

- Case 3: $C$ is in the first or last spot. Then, we have two choices for the position of $C$. Without loss of generality, let it be in the first spot. Either $A$ or $E$ is in the second spot, giving us two choices. Without loss of generality, let it be $A$. Then, if $D$ is in the third spot, the positions of $B$ and $E$ are fixed. If $E$ is in third spot, the positions of $B$ and $D$ are fixed, so we have a total of $2 \times 2 \times (1+1) = 8$ options for this case.

Hence, we have a total of $2 + 4 + 8 = 14$ possibilities.