Finding all descendants of node

[Ram Rachum]

Hi guys,

I'm a newb at graph theory and networkx, so please excuse my ignorance.

Assume I have a tree in networkx. How do I find all its descendants recursively? (i.e., all its children, and its children's children, etc.)?

Followup question: How do I find all its children recursively that are leaves (i.e. have no children.)

Thanks,

Ram.

[Patrick Dolan]

Ram, 

I think you this set of functions. Depth first search and breadth first search: http://networkx.lanl.gov/reference/algorithms.traversal.html

Patrick Dolan

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[Ram Rachum]

Thanks! I'll give it a try. 

[Ram Rachum]

I checked it out, and I couldn't figure out which of these functions finds all children, recursively. (Including children of children, and children of children of children, etc.) Which is it? 

Thanks,

Ram.

[Ram Rachum]

My mistake, I see that dfs_preorder_nodes provides this. Thanks!

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[Ram Rachum]

Though I wonder, why aren't there bfs versions of these two functions?

[Aric Hagberg]

I suppose nobody ever wrote a bfs_nodes() function since it is one
line of python:
In [1]: import networkx as nx

In [2]: G = nx.path_graph(4)

In [3]: [v for u,v in nx.bfs_edges(G,0)]
Out[3]: [1, 2, 3]

But we could consider adding that if that is a commonly used expression.

Aric