我没有找到距离矩阵的解决方案，但似乎我忽略了使用树的query方法的最明显的解决方案。
所以为了找到第一近邻之间的最大距离，我做了（向量是一个形状为（N，3）的numpy数组）：

tree = cKDTree(vectors, leaf_size)
# get the indexes of the first nearest neighbor of each vertex
# we use k=2 because k=1 are the points themselves with distance 0
nn1 = tree.query(vectors, k=2)[1][:,1]
# get the vectors corresponding to those indexes. Basically this is "vectors" sorted by
# first nearest neighbor of each point in "vectors".
nn1_vec = vectors[nn1]
# the distance between each point and its first nearest neighbor
nn_dist = np.sqrt(np.sum((vectors - nn1_vec)**2, axis=1))
# maximum distance
return np.max(nn_dist)

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如果有人发现这个以后（像我）。这里有一个稍微简单的版本。感谢DIN14970做所有的研究。
结果是.query返回的是距离（最初询问时可能不是这种情况）。不需要计算它们。

tree = cKDTree(vectors, leaf_size)
# get the indexes of the first nearest neighbor of each vertex
# we use k=2 because k=1 are the points themselves with distance 0
nn1_distance = tree.query(vectors, k=2)[1][0][:,1]
# maximum distance
return np.max(nn1_distance)

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