如何在Networkx中手动实施DFS

Question

I would like to implement by hand a simple DFS preorder traversal of a tree in networkx . As a toy example, here is a simple tree:

import matplotlib.pyplot as plt
import networkx as nx
T = nx.generators.balanced_tree(2, 3)
nx.draw_networkx(T)
plt.show()

As this is a binary tree, I should in principle be able to perform a preorder traversal with something like:

def preOrder(root):
    if root:
        print(root.data)
        preOrder(root.left)
        preOrder(root.right)

However, I can't see how to do this in networkx. How do I find the root and what is the networkx version of node.left and node.right?

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Update 1

In networkx there is list(nx.dfs_preorder_nodes(T, source=r, depth_limit=1)) which treats the tree as an undirected graph and finds all the children from r which it treats as the root. I don't see how to use this, for example, to mimic node.left and node.right to perform the preorder traversal. It will for example give [0,3,4] if r = 1 .

Answer 1

Networkx allows you varied ways to traverse nodes of a graph. In your case you could use dfs_preorder_nodes and limit depth to 1 :

>>> G = nx.path_graph(5)
>>> list(nx.dfs_preorder_nodes(G, source=0))
[0, 1, 2, 3, 4]
>>> list(nx.dfs_preorder_nodes(G, source=0, depth_limit=1))
[0, 1]

Note the API:

If a source is not specified then a source is chosen arbitrarily and repeatedly until all components in the graph are searched.