已验证A*扩展节点更少，哪些场景下BFS、DFS比A*更高效？

Great question! It's totally reasonable to assume A* is the go-to since it's optimal and efficient with a good heuristic, but there are concrete scenarios where BFS or DFS are more practical or faster. Let's dive into them:

1. Poor or Non-Existent Heuristic Functions #

A*'s superpower relies entirely on a good, admissible heuristic (one that never overestimates the cost to reach the goal). If you can't design such a heuristic—maybe the problem domain has no obvious way to estimate distance to the goal, or your heuristic is wildly inaccurate—A* loses its edge.

For example, in a completely unknown, randomly generated state space where you have zero domain knowledge, guessing a heuristic might lead A* to expand more nodes than a straightforward BFS or DFS. In cases where the heuristic is inadmissible (overestimates), A* might even fail to find the optimal path, making BFS (for uniform cost) or DFS (for any solution) better choices.

2. Memory-Limited Environments #

A* needs to maintain two key data structures: an open list (nodes to explore) and a closed list (nodes already explored). For large state spaces, these lists can grow exponentially, eating up massive amounts of memory.

DFS, by contrast, only needs to keep track of the current path via a stack—its memory usage is linear with the depth of the current path, which is way more efficient for deep state spaces. This makes DFS ideal for memory-constrained systems like embedded devices or low-resource environments where A* would crash due to memory exhaustion.

BFS also has lower memory overhead than A* in some cases, especially when the goal is in the shallow layers of the state space, as its queue only holds nodes from the current and next layers.

3. Any Valid Solution Works (No Need for Optimality) #

If you don't care about finding the optimal path—you just need any solution—DFS can often find an answer faster than A*. For example, in puzzle-solving where you just need one valid solution (not the shortest), DFS might dive straight down a path and hit the goal before A* finishes evaluating lower-cost paths.

Similarly, if the goal is extremely close to the start state, BFS might reach it faster than A* because it doesn't waste time calculating heuristic values for each node—it just explores layers in order.

4. Small, Simple State Spaces #

When the total number of possible states is tiny (think dozens or hundreds of nodes), the overhead of calculating A*'s f(n) = g(n) + h(n) for each node becomes a burden. BFS or DFS can traverse the entire state space in less time than A* spends computing heuristic values.

In these cases, brute-force traversal is simpler to implement and faster to run, since the cost of the heuristic calculation outweighs any benefit A* would provide.

5. Uniform Cost Paths (BFS-Specific Edge Case) #

If every edge in your graph has the same cost, BFS is already guaranteed to find the shortest path—just like A* with a heuristic h(n) = 0. But BFS is simpler to code, has no heuristic calculation overhead, and avoids the bookkeeping of A*'s open/closed lists. In this scenario, BFS is just as optimal as A* but more efficient in practice.

内容的提问来源于stack exchange，提问作者Marjan