From 080d216ffd901bac776c77dba23bf8fd7adf8c9a Mon Sep 17 00:00:00 2001
From: Neko Box Coder <nekoboxcoder@gmail.com>
Date: Sun, 30 Mar 2025 14:17:24 +0100
Subject: [PATCH 1/2] Fixing chained parents by flattening the tree on unsplit

---
 internal/views/splits.go | 49 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 49 insertions(+)

diff --git a/internal/views/splits.go b/internal/views/splits.go
index b28a6c09..1e4e96b4 100644
--- a/internal/views/splits.go
+++ b/internal/views/splits.go
@@ -479,9 +479,58 @@ func (n *Node) Unsplit() bool {
 	if n.parent.IsLeaf() {
 		return n.parent.Unsplit()
 	}
+
+	n.parent.flatten()
 	return true
 }
 
+// flattens the tree by removing unnecessary intermediate parents that have only one child
+// and handles the side effect of it
+func (n *Node) flatten() {
+	if n.parent == nil || len(n.children) != 1 {
+		return
+	}
+
+	ind := 0
+	for i, c := range n.parent.children {
+		if c.id == n.id {
+			ind = i
+		}
+	}
+
+	// Replace current node with its child node to remove chained parent
+	successor := n.children[0]
+	n.parent.children[ind] = successor
+	successor.parent = n.parent
+
+	// Maintain the tree in a consistent state: any child node's kind (horiz vs vert)
+	// should be the opposite of its parent's kind.
+	if successor.IsLeaf() {
+		successor.Kind = n.Kind
+	} else {
+		// If the successor node has children, that means it is a chained parent as well.
+		// Therefore it can be replaced by its own children.
+		origsize := len(n.parent.children)
+
+		// Let's say we have 5 children and want to replace [2] with its children [a] [b] [c]
+		// [0] [1] [2] [3] [4] --> [0] [1] [a] [b] [c] [3] [4]
+		// insertcount will be `3 - 1 = 2` in this case
+		insertcount := len(successor.children) - 1
+
+		n.parent.children = append(n.parent.children, make([]*Node, insertcount)...)
+		copy(n.parent.children[ind+insertcount+1:], n.parent.children[ind+1:origsize])
+		copy(n.parent.children[ind:], successor.children)
+
+		for i := 0; i < len(successor.children); i++ {
+			n.parent.children[ind+i].parent = n.parent
+		}
+	}
+
+	// Update propW and propH since the parent of the children has been updated,
+	// so the children have new siblings
+	n.parent.markSizes()
+}
+
 // String returns the string form of the node and all children (used for debugging)
 func (n *Node) String() string {
 	var strf func(n *Node, ident int) string

From 8e7089993dfdd01c70fd7c6b8d9ef7792370076a Mon Sep 17 00:00:00 2001
From: Neko Box Coder <nekoboxcoder@gmail.com>
Date: Sat, 5 Apr 2025 23:34:59 +0100
Subject: [PATCH 2/2] Simplifying unsplit logic

---
 internal/views/splits.go | 10 +++-------
 1 file changed, 3 insertions(+), 7 deletions(-)

diff --git a/internal/views/splits.go b/internal/views/splits.go
index 1e4e96b4..b2a20873 100644
--- a/internal/views/splits.go
+++ b/internal/views/splits.go
@@ -439,11 +439,12 @@ func (n *Node) VSplit(right bool) uint64 {
 }
 
 // unsplits the child of a split
-func (n *Node) unsplit(i int, h bool) {
+func (n *Node) unsplit(i int) {
 	copy(n.children[i:], n.children[i+1:])
 	n.children[len(n.children)-1] = nil
 	n.children = n.children[:len(n.children)-1]
 
+	h := n.Kind == STVert
 	nonrs, numr := n.getResizeInfo(h)
 	if numr == 0 {
 		// This means that this was the last child
@@ -470,12 +471,7 @@ func (n *Node) Unsplit() bool {
 			ind = i
 		}
 	}
-	if n.parent.Kind == STVert {
-		n.parent.unsplit(ind, true)
-	} else {
-		n.parent.unsplit(ind, false)
-	}
-
+	n.parent.unsplit(ind)
 	if n.parent.IsLeaf() {
 		return n.parent.Unsplit()
 	}