otg/core/src/types.rs

@@ -241,7 +241,7 @@ pub struct Tree<T> {
 
 #[derive(Debug)]
 pub struct Node<T> {
-    id: usize,
+    pub id: usize,
     node: T,
     parent: Option<usize>,
     depth: usize,
@@ -355,7 +355,7 @@ impl<T> Tree<T> {
         width
     }
 
-    fn bfs_iter<'a>(&'a self) -> BFSIter<T> {
+    pub fn bfs_iter<'a>(&'a self) -> BFSIter<T> {
         let mut queue = VecDeque::new();
         queue.push_back(&self.nodes[0]);
         BFSIter { tree: self, queue }

otg/gtk/src/components/review_tree.rs

@@ -66,8 +66,29 @@ impl ReviewTree {
         s
     }
 
-    pub fn redraw(&self, _ctx: &Context, _width: i32, _height: i32) {
+    pub fn redraw(&self, ctx: &Context, _width: i32, _height: i32) {
-        // Implement the tree-drawing algorithm here
+        println!("redraw");
+        let tree: &Option<Tree<Uuid>> = &self.imp().tree.borrow();
+        match tree {
+            Some(ref tree) => {
+                for node in tree.bfs_iter() {
+                    // draw a circle given the coordinates of the nodes
+                    // I don't know the indent. How do I keep track of that? Do I track the position of
+                    // the parent? do I need to just make it more intrinsically a part of the position
+                    // code?
+                    ctx.set_source_rgb(0.7, 0.7, 0.7);
+                    let (row, column) = tree.position(0, node.id);
+                    println!("[{}] {} x {}", node.id, row, column);
+                    let y = (row as f64) * 20. + 10.;
+                    let x = (column as f64) * 20. + 10.;
+                    ctx.arc(x, y, 5., 0., 2. * std::f64::consts::PI);
+                    let _ = ctx.stroke();
+                }
+            }
+            None => {
+                // if there is no tree present, then there's nothing to draw!
+            }
+        }
     }
 }
 
@@ -110,43 +131,6 @@ struct Tree {
 }
 */
 
-// Given a node, do a postorder traversal to figure out the width of the node based on all of its
-// children. This is equivalent to the widest of all of its children at all depths.
-//
-// There are some collapse rules that I could take into account here, but that I haven't figured
-// out yet. If two nodes are side by side, and one of them has some wide children but the other has
-// no children, then they are effectively the same width. The second node only needs to be moved
-// out if it has children that would overlap the children of the first node.
-//
-// My algorithm right now is likely to generate unnecessarily wide trees in a complex game review.
-#[allow(dead_code)]
-fn node_width(node: &GameNode) -> usize {
-    let children: &Vec<GameNode> = match node {
-        GameNode::MoveNode(mn) => &mn.children,
-        GameNode::SetupNode(sn) => &sn.children,
-    };
-
-    if children.is_empty() {
-        return 1;
-    }
-
-    // If there is more than one child, run node_width on each one and add them together.
-    children
-        .iter()
-        .fold(0, |acc, child| acc + node_width(child))
-}
-
-// Since I know the width of a node, now I want to figure out its placement in the larger scheme of
-// things.
-//
-// One thought I have is that I could just develop a grid virtually and start placing nodes.
-// Whenever I notice a collision, I can just move the node over. But I'd like to see if I can be a
-// bit smarter than doing it as just a vec into which I place things, as though it's a game board.
-// So, given a game node, I want to figure out it's position along the X axis.
-//
-// Just having the node is greatly insufficient. I can get better results if I'm calculating the
-// position of its children.
-
 #[cfg(test)]
 mod test {
     use super::*;