Example 2 Let ABCDEF be a regular hexagon. A frog starts at vertex $A$, and each time it can randomly jump to one of the two adjacent vertices. If it reaches point $D$ within 5 jumps, it stops jumping; if it cannot reach point $D$ within 5 jumps, it also stops after 5 jumps. Then, the number of different possible jumping sequences from the start to the stop is $\qquad$ kinds.
Solution: As shown in Figure 1, the frog cannot reach point $D$ by jumping 1 time, 2 times, or 4 times. Therefore, the frog's jumping methods are only in the following two cases:
(1) The frog jumps 3 times to reach point $D$, with two jumping methods: $A B C D$ and $A F E D$.
(2) The frog stops after a total of 5 jumps. In this case, the first 3 jumps must not reach $D$, but can only reach $B$ or $F$. There are 6 jumping methods: $A F E F$, $A F A F$, $A B A F$, $A B C B$, $A B A B$, and $A F A B$. The subsequent 2 jumps each have 4 methods, for example, from $F$, there are $F E F$, $F E D$, $F A F$, and $F A B$, a total of 4 methods. Therefore, these 5 jumps have $6 \times 4 = 24$ different methods.
Thus, there are a total of $2 + 24 = 26$ different jumping methods.