skbio.tree.TreeNode.is_tip# TreeNode.is_tip()[source]# Return True if the current node has no children. Returns: boolTrue if the node is a tip See also is_root has_children Examples >>> from skbio import TreeNode >>> tree = TreeNode.read(["((a,b)c);"]) >>> print(tree.is_tip()) False >>> print(tree.find('a').is_tip()) True