diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md
index 94eb405b..f8109f85 100644
--- a/problems/0106.从中序与后序遍历序列构造二叉树.md
+++ b/problems/0106.从中序与后序遍历序列构造二叉树.md
@@ -1149,6 +1149,52 @@ object Solution {
 }
 ```
 
+## rust
+
+106 从中序与后序遍历序列构造二叉树
+
+```rust
+use std::cell::RefCell;
+use std::rc::Rc;
+impl Solution {
+    pub fn build_tree(inorder: Vec<i32>, postorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
+        if inorder.is_empty() {
+            return None;
+        }
+        let mut postorder = postorder;
+        let root = postorder.pop().unwrap();
+        let index = inorder.iter().position(|&x| x == root).unwrap();
+        let mut root = TreeNode::new(root);
+        root.left = Self::build_tree(inorder[..index].to_vec(), postorder[..index].to_vec());
+        root.right = Self::build_tree(inorder[index + 1..].to_vec(), postorder[index..].to_vec());
+        Some(Rc::new(RefCell::new(root)))
+    }
+}
+```
+
+105 从前序与中序遍历序列构造二叉树
+
+```rust
+use std::cell::RefCell;
+use std::rc::Rc;
+impl Solution {
+    pub fn build_tree(preorder: Vec<i32>, inorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
+        if preorder.is_empty() {
+            return None;
+        }
+        let root = preorder[0];
+        let index = inorder.iter().position(|&x| x == root).unwrap();
+        let mut root = TreeNode::new(root);
+        root.left = Self::build_tree(preorder[1..index + 1].to_vec(), inorder[0..index].to_vec());
+        root.right = Self::build_tree(
+            preorder[index + 1..].to_vec(),
+            inorder[index + 1..].to_vec(),
+        );
+        Some(Rc::new(RefCell::new(root)))
+    }
+}
+```
+
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
   <img src="../pics/网站星球宣传海报.jpg" width="1000"/>
diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md
index ee19e53d..8f4ca2d4 100644
--- a/problems/背包理论基础01背包-1.md
+++ b/problems/背包理论基础01背包-1.md
@@ -271,39 +271,64 @@ int main() {
 ### java 
 
 ```java
+public class BagProblem {
     public static void main(String[] args) {
-        int[] weight = {1, 3, 4};
-        int[] value = {15, 20, 30};
-        int bagsize = 4;
-        testweightbagproblem(weight, value, bagsize);
+        int[] weight = {1,3,4};
+        int[] value = {15,20,30};
+        int bagSize = 4;
+        testWeightBagProblem(weight,value,bagSize);
     }
 
-    public static void testweightbagproblem(int[] weight, int[] value, int bagsize){
-        int wlen = weight.length, value0 = 0;
-        //定义dp数组：dp[i][j]表示背包容量为j时，前i个物品能获得的最大价值
-        int[][] dp = new int[wlen + 1][bagsize + 1];
-        //初始化：背包容量为0时，能获得的价值都为0
-        for (int i = 0; i <= wlen; i++){
-            dp[i][0] = value0;
+    /**
+     * 动态规划获得结果
+     * @param weight  物品的重量
+     * @param value   物品的价值
+     * @param bagSize 背包的容量
+     */
+    public static void testWeightBagProblem(int[] weight, int[] value, int bagSize){
+
+        // 创建dp数组
+        int goods = weight.length;  // 获取物品的数量
+        int[][] dp = new int[goods][bagSize + 1];
+
+        // 初始化dp数组
+        // 创建数组后，其中默认的值就是0
+        for (int j = weight[0]; j <= bagSize; j++) {
+            dp[0][j] = value[0];
         }
-        //遍历顺序：先遍历物品，再遍历背包容量
-        for (int i = 1; i <= wlen; i++){
-            for (int j = 1; j <= bagsize; j++){
-                if (j < weight[i - 1]){
-                    dp[i][j] = dp[i - 1][j];
-                }else{
-                    dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1]);
+
+        // 填充dp数组
+        for (int i = 1; i < weight.length; i++) {
+            for (int j = 1; j <= bagSize; j++) {
+                if (j < weight[i]) {
+                    /**
+                     * 当前背包的容量都没有当前物品i大的时候，是不放物品i的
+                     * 那么前i-1个物品能放下的最大价值就是当前情况的最大价值
+                     */
+                    dp[i][j] = dp[i-1][j];
+                } else {
+                    /**
+                     * 当前背包的容量可以放下物品i
+                     * 那么此时分两种情况：
+                     *    1、不放物品i
+                     *    2、放物品i
+                     * 比较这两种情况下，哪种背包中物品的最大价值最大
+                     */
+                    dp[i][j] = Math.max(dp[i-1][j] , dp[i-1][j-weight[i]] + value[i]);
                 }
             }
         }
-        //打印dp数组
-        for (int i = 0; i <= wlen; i++){
-            for (int j = 0; j <= bagsize; j++){
-                System.out.print(dp[i][j] + " ");
+
+        // 打印dp数组
+        for (int i = 0; i < goods; i++) {
+            for (int j = 0; j <= bagSize; j++) {
+                System.out.print(dp[i][j] + "\t");
             }
-            System.out.print("\n");
+            System.out.println("\n");
         }
     }
+}
+
 ```
 
 ### python