skbio.core.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 skbio.core.tree import TreeNode
>>> tree = TreeNode.from_newick("((a,b)c);")
>>> print tree.is_tip()
False
>>> print tree.find('a').is_tip()
True