Make insert new key as red node with two empty sub-trees clear

liuxinyu95 · Jan 21, 2022 · 25bff64 · 25bff64
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diff --git a/datastruct/tree/red-black-tree/rbtree-en.tex b/datastruct/tree/red-black-tree/rbtree-en.tex
@@ -275,7 +275,7 @@ \section{Insert}
 \end{array}
 \ee
 
-If the tree is empty, we create a red leaf of $k$; otherwise, let the sub-trees and the key be $l$, $r$, $k'$, we compare $k$ and $k'$, then recursively insert $k$ to a sub-tree. After that, we call $balance$ to fix the coloring, then force the root to be black finally.
+If the tree is empty, we create a red node of $k$ with two empty sub-trees; otherwise, let the sub-trees and the key be $l$, $r$, $k'$, we compare $k$ and $k'$, then recursively insert $k$ to a sub-tree. After that, we call $balance$ to fix the coloring, then force the root to be black finally.
 
 \be
 makeBlack\ (\mathcal{C}, l, k, r) = (\mathcal{B}, l, k, r)

diff --git a/datastruct/tree/red-black-tree/rbtree-zh-cn.tex b/datastruct/tree/red-black-tree/rbtree-zh-cn.tex
@@ -278,7 +278,7 @@ \section{插入}
 \end{array}
 \ee
 
-如果树为空，我们为$k$创建一个红色叶子节点；否则，令树的左右分支和值分别为$l$、$r$、$k'$。比较$k$和$k'$的大小，递归地将$k$插入到子树中。然后用$balance$修复平衡性。最后强制把根节点染成黑色。
+如果树为空，我们为$k$创建一个红色节点，它的两个分枝都为空；否则，令树的左右分支和值分别为$l$、$r$、$k'$。比较$k$和$k'$的大小，递归地将$k$插入到子树中。然后用$balance$修复平衡性。最后强制把根节点染成黑色。
 
 \be
 makeBlack\ (\mathcal{C}, l, k, r) = (\mathcal{B}, l, k, r)