skbio.tree.TreeNode.is_tip¶

TreeNode.is_tip()[source]¶

Returns True if the current node has no children.

Returns:	bool True if the node is a tip

See also

is_root, has_children

Examples

>>> from six import StringIO
>>> from skbio import TreeNode
>>> tree = TreeNode.read(StringIO("((a,b)c);"))
>>> print(tree.is_tip())
False
>>> print(tree.find('a').is_tip())
True