simplify: use direct node.children check instead of has_children() method

Even simpler approach:
- Replace node.has_children() with direct if node.children:
- Remove unnecessary has_children() methods from all classes
- Pythonic and direct - empty lists are falsy, non-empty are truthy
- Less code, same functionality

This is the most straightforward way to check for children in Python.

Author: cursoragent

src/gitingest/output_formatter.py

@@ -156,7 +156,7 @@ def _create_tree_structure(
     display_name = node.get_display_name()
     tree_str += f"{prefix}{current_prefix}{display_name}\n"
 
-    if node.has_children():
+    if node.children:
         prefix += "    " if is_last else "│   "
         for i, child in enumerate(node.children):
             tree_str += _create_tree_structure(query, node=child, prefix=prefix, is_last=i == len(node.children) - 1)

src/gitingest/schemas/filesystem.py

@@ -112,9 +112,7 @@ def get_display_name(self) -> str:
         """Get display name for tree view. Override in subclasses."""
         return self.name
 
-    def has_children(self) -> bool:
-        """Return whether this node has children to display."""
-        return False
+
 
     @property
     def content(self) -> str:
@@ -167,9 +165,7 @@ def get_display_name(self) -> str:
         """Directories get a trailing slash."""
         return self.name + "/"
 
-    def has_children(self) -> bool:
-        """Directories have children if the list is not empty."""
-        return bool(self.children)
+
 
     def render_tree(self, prefix: str = "", *, is_last: bool = True) -> list[str]:
         """Render the tree representation of this directory."""