skbio.tree.TreeNode.is_tip¶
-
TreeNode.
is_tip
()[source]¶ Returns True if the current node has no children.
State: Experimental as of 0.4.0.
- Returns
True if the node is a tip
- Return type
See also
Examples
>>> from skbio import TreeNode >>> tree = TreeNode.read(["((a,b)c);"]) >>> print(tree.is_tip()) False >>> print(tree.find('a').is_tip()) True