diff --git a/codes/python/chapter_divide_and_conquer/build_tree.py b/codes/python/chapter_divide_and_conquer/build_tree.py
index 85266162c..b8b2db293 100644
--- a/codes/python/chapter_divide_and_conquer/build_tree.py
+++ b/codes/python/chapter_divide_and_conquer/build_tree.py
@@ -18,7 +18,7 @@ def dfs(
     l: int,
     r: int,
 ) -> TreeNode | None:
-    """构建二叉树 DFS"""
+    """构建二叉树：分治"""
     # 子树区间为空时终止
     if r - l < 0:
         return None
@@ -26,9 +26,9 @@ def dfs(
     root = TreeNode(preorder[i])
     # 查询 m ，从而划分左右子树
     m = hmap[preorder[i]]
-    # 递归构建左子树
+    # 子问题：构建左子树
     root.left = dfs(preorder, inorder, hmap, i + 1, l, m - 1)
-    # 递归构建右子树
+    # 子问题：构建右子树
     root.right = dfs(preorder, inorder, hmap, i + 1 + m - l, m + 1, r)
     # 返回根节点
     return root
diff --git a/docs/chapter_data_structure/summary.md b/docs/chapter_data_structure/summary.md
index a01f6332b..74cdc549e 100644
--- a/docs/chapter_data_structure/summary.md
+++ b/docs/chapter_data_structure/summary.md
@@ -22,6 +22,6 @@
 
     哈希表底层是数组，而为了解决哈希冲突，我们可能会使用“拉链法”（后续散列表章节会讲）。在拉链法中，数组中每个地址（桶）指向一个链表；当这个链表长度超过一定阈值时，又可能被转化为树（通常为红黑树）。因此，哈希表可能同时包含线性（数组、链表）和非线性（树）数据结构。
 
-!!! question "char 类型的长度是 1 bytes 吗？"
+!!! question "char 类型的长度是 1 byte 吗？"
 
-    这个与编程语言采用的编码方法有关。例如，Java, JS, TS, C# 都采用 UTF-16 编码（保存 Unicode 码点），因此 char 类型的长度为 2 bytes 。
+    char 类型的长度由编程语言采用的编码方法决定。例如，Java, JS, TS, C# 都采用 UTF-16 编码（保存 Unicode 码点），因此 char 类型的长度为 2 bytes 。
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/build_tree_division_pointers.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/build_tree_division_pointers.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/build_tree_division_pointers.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/build_tree_division_pointers.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/build_tree_example.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/build_tree_example.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/build_tree_example.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/build_tree_example.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/build_tree_preorder_inorder_division.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/build_tree_preorder_inorder_division.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/build_tree_preorder_inorder_division.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/build_tree_preorder_inorder_division.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step1.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step1.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step1.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step1.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step10.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step10.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step10.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step10.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step2.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step2.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step2.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step2.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step3.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step3.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step3.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step3.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step4.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step4.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step4.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step4.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step5.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step5.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step5.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step5.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step6.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step6.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step6.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step6.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step7.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step7.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step7.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step7.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step8.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step8.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step8.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step8.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step9.png b/docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step9.png
similarity index 100%
rename from docs/chapter_divide_and_conquer/build_binary_tree.assets/built_tree_step9.png
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.assets/built_tree_step9.png
diff --git a/docs/chapter_divide_and_conquer/build_binary_tree.md b/docs/chapter_divide_and_conquer/build_binary_tree_problem.md
similarity index 83%
rename from docs/chapter_divide_and_conquer/build_binary_tree.md
rename to docs/chapter_divide_and_conquer/build_binary_tree_problem.md
index 63f46bb8e..872ffeeab 100644
--- a/docs/chapter_divide_and_conquer/build_binary_tree.md
+++ b/docs/chapter_divide_and_conquer/build_binary_tree_problem.md
@@ -4,7 +4,7 @@
 
     给定一个二叉树的前序遍历 `preorder` 和中序遍历 `inorder` ，请从中构建二叉树，返回二叉树的根节点。
 
-![构建二叉树的示例数据](build_binary_tree.assets/build_tree_example.png)
+![构建二叉树的示例数据](build_binary_tree_problem.assets/build_tree_example.png)
 
 原问题定义为从 `preorder` 和 `inorder` 构建二叉树。我们首先从分治的角度分析这道题：
 
@@ -25,7 +25,7 @@
 2. 查找根节点在 `inorder` 中的索引，基于该索引可将 `inorder` 划分为 `[ 9 | 3 ｜ 1 2 7 ]` ；
 3. 根据 `inorder` 划分结果，可得左子树和右子树分别有 1 个和 3 个节点，从而可将 `preorder` 划分为 `[ 3 | 9 | 2 1 7 ]` ；
 
-![在前序和中序遍历中划分子树](build_binary_tree.assets/build_tree_preorder_inorder_division.png)
+![在前序和中序遍历中划分子树](build_binary_tree_problem.assets/build_tree_preorder_inorder_division.png)
 
 至此，**我们已经推导出根节点、左子树、右子树在 `preorder` 和 `inorder` 中的索引区间**。而为了描述这些索引区间，我们需要借助几个指针变量：
 
@@ -47,7 +47,7 @@
 
 请注意，右子树根节点索引中的 $(m-l)$ 的含义是“左子树的节点数量”，建议配合下图理解。
 
-![根节点和左右子树的索引区间表示](build_binary_tree.assets/build_tree_division_pointers.png)
+![根节点和左右子树的索引区间表示](build_binary_tree_problem.assets/build_tree_division_pointers.png)
 
 接下来就可以实现代码了。为了提升查询 $m$ 的效率，我们借助一个哈希表 `hmap` 来存储 `inorder` 列表元素到索引的映射。
 
@@ -142,34 +142,34 @@
 下图展示了构建二叉树的递归过程，各个节点是在向下“递”的过程中建立的，而各条边是在向上“归”的过程中建立的。
 
 === "<1>"
-    ![built_tree_step1](build_binary_tree.assets/built_tree_step1.png)
+    ![构建二叉树的递归过程](build_binary_tree_problem.assets/built_tree_step1.png)
 
 === "<2>"
-    ![built_tree_step2](build_binary_tree.assets/built_tree_step2.png)
+    ![built_tree_step2](build_binary_tree_problem.assets/built_tree_step2.png)
 
 === "<3>"
-    ![built_tree_step3](build_binary_tree.assets/built_tree_step3.png)
+    ![built_tree_step3](build_binary_tree_problem.assets/built_tree_step3.png)
 
 === "<4>"
-    ![built_tree_step4](build_binary_tree.assets/built_tree_step4.png)
+    ![built_tree_step4](build_binary_tree_problem.assets/built_tree_step4.png)
 
 === "<5>"
-    ![built_tree_step5](build_binary_tree.assets/built_tree_step5.png)
+    ![built_tree_step5](build_binary_tree_problem.assets/built_tree_step5.png)
 
 === "<6>"
-    ![built_tree_step6](build_binary_tree.assets/built_tree_step6.png)
+    ![built_tree_step6](build_binary_tree_problem.assets/built_tree_step6.png)
 
 === "<7>"
-    ![built_tree_step7](build_binary_tree.assets/built_tree_step7.png)
+    ![built_tree_step7](build_binary_tree_problem.assets/built_tree_step7.png)
 
 === "<8>"
-    ![built_tree_step8](build_binary_tree.assets/built_tree_step8.png)
+    ![built_tree_step8](build_binary_tree_problem.assets/built_tree_step8.png)
 
 === "<9>"
-    ![built_tree_step9](build_binary_tree.assets/built_tree_step9.png)
+    ![built_tree_step9](build_binary_tree_problem.assets/built_tree_step9.png)
 
 === "<10>"
-    ![built_tree_step10](build_binary_tree.assets/built_tree_step10.png)
+    ![built_tree_step10](build_binary_tree_problem.assets/built_tree_step10.png)
 
 设树的节点数量为 $n$ ，初始化每一个节点（执行一个递归函数 `dfs()` ）使用 $O(1)$ 时间。**因此总体时间复杂度为 $O(n)$** 。
 
diff --git a/docs/chapter_dynamic_programming/intro_to_dynamic_programming.md b/docs/chapter_dynamic_programming/intro_to_dynamic_programming.md
index 68ecaf738..5fbf49d26 100644
--- a/docs/chapter_dynamic_programming/intro_to_dynamic_programming.md
+++ b/docs/chapter_dynamic_programming/intro_to_dynamic_programming.md
@@ -476,5 +476,5 @@ $$
 总的看来，**子问题分解是一种通用的算法思路，在分治、动态规划、回溯中各有特点**：
 
 - 分治算法将原问题划分为几个独立的子问题，然后递归解决子问题，最后合并子问题的解得到原问题的解。例如，归并排序将长数组不断划分为两个短子数组，再将排序好的子数组合并为排序好的长数组。
-- 动态规划也是将原问题分解为多个子问题，但与分治算法的主要区别是，**动态规划中的子问题往往不是相互独立的**，原问题的解依赖于子问题的解，而子问题的解又依赖于更小的子问题的解。因此，动态规划通常会引入记忆化，保存已经解决的子问题的解，避免重复计算。
-- 回溯算法在尝试和回退中穷举所有可能的解，并通过剪枝避免不必要的搜索分支。原问题的解由一系列决策步骤构成，我们可以将每个决策步骤之后的剩余问题看作为一个子问题。
+- 动态规划也是将原问题分解为多个子问题，但与分治算法的主要区别是，**动态规划中的子问题往往不是相互独立的**，原问题的解依赖于子问题的解，而子问题的解又依赖于更小的子问题的解。
+- 回溯算法在尝试和回退中穷举所有可能的解，并通过剪枝避免不必要的搜索分支。原问题的解由一系列决策步骤构成，我们可以将每个决策步骤之前的子序列看作为一个子问题。
diff --git a/docs/chapter_sorting/bucket_sort.md b/docs/chapter_sorting/bucket_sort.md
index bcb40b7c8..22ece0854 100644
--- a/docs/chapter_sorting/bucket_sort.md
+++ b/docs/chapter_sorting/bucket_sort.md
@@ -86,7 +86,7 @@
 
 ## 算法特性
 
-- **时间复杂度 $O(n + k)$** ：假设元素在各个桶内平均分布，那么每个桶内的元素数量为 $\frac{n}{k}$ 。假设排序单个桶使用 $O(\frac{n}{k} \log\frac{n}{k})$ 时间，则排序所有桶使用 $O(n \log\frac{n}{k})$ 时间。**当桶数量 $k$ 比较大时，时间复杂度则趋向于 $O(n)$** 。合并结果时需要遍历 $n$ 个桶，花费 $O(k)$ 时间。
+- **时间复杂度 $O(n + k)$** ：假设元素在各个桶内平均分布，那么每个桶内的元素数量为 $\frac{n}{k}$ 。假设排序单个桶使用 $O(\frac{n}{k} \log\frac{n}{k})$ 时间，则排序所有桶使用 $O(n \log\frac{n}{k})$ 时间。**当桶数量 $k$ 比较大时，时间复杂度则趋向于 $O(n)$** 。合并结果时需要遍历所有桶和元素，花费 $O(n + k)$ 时间。
 - **自适应排序**：在最坏情况下，所有数据被分配到一个桶中，且排序该桶使用 $O(n^2)$ 时间。
 - **空间复杂度 $O(n + k)$ 、非原地排序** ：需要借助 $k$ 个桶和总共 $n$ 个元素的额外空间。
 - 桶排序是否稳定取决于排序桶内元素的算法是否稳定。