diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md
index e3fb0fb5..bf1e173e 100644
--- a/problems/0001.两数之和.md
+++ b/problems/0001.两数之和.md
@@ -163,7 +163,7 @@ class Solution:
         for index, value in enumerate(nums):  
             if target - value in records:   # 遍历当前元素，并在map中寻找是否有匹配的key
                 return [records[target- value], index]
-            records[value] = index    # 遍历当前元素，并在map中寻找是否有匹配的key
+            records[value] = index    # 如果没找到匹配对，就把访问过的元素和下标加入到map中
         return []
 ```
 （版本二）使用集合
diff --git a/problems/0077.组合.md b/problems/0077.组合.md
index 7d71b517..8d448739 100644
--- a/problems/0077.组合.md
+++ b/problems/0077.组合.md
@@ -343,8 +343,32 @@ public:
 ## 其他语言版本
 
 
-### Java
 
+### Java：
+未剪枝优化
+```java
+class Solution {
+    List<List<Integer>> result= new ArrayList<>();
+    LinkedList<Integer> path = new LinkedList<>();
+    public List<List<Integer>> combine(int n, int k) {
+        backtracking(n,k,1);
+        return result;
+    }
+
+    public void backtracking(int n,int k,int startIndex){
+        if (path.size() == k){
+            result.add(new ArrayList<>(path));
+            return;
+        }
+        for (int i =startIndex;i<=n;i++){
+            path.add(i);
+            backtracking(n,k,i+1);
+            path.removeLast();
+        }
+    }
+}
+```
+剪枝优化：
 ```java
 class Solution {
     List<List<Integer>> result = new ArrayList<>();
diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md
index 13599fd8..18e5cb4c 100644
--- a/problems/0450.删除二叉搜索树中的节点.md
+++ b/problems/0450.删除二叉搜索树中的节点.md
@@ -268,6 +268,34 @@ public:
 
 
 ### Java 
+```java
+// 解法1(最好理解的版本)
+class Solution {
+  public TreeNode deleteNode(TreeNode root, int key) {
+    if (root == null) return root;
+    if (root.val == key) {
+      if (root.left == null) {
+        return root.right;
+      } else if (root.right == null) {
+        return root.left;
+      } else {
+        TreeNode cur = root.right;
+        while (cur.left != null) {
+          cur = cur.left;
+        }
+        cur.left = root.left;
+        root = root.right;
+        return root;
+      }
+    }
+    if (root.val > key) root.left = deleteNode(root.left, key);
+    if (root.val < key) root.right = deleteNode(root.right, key);
+    return root;
+  }
+}
+```
+
+
 ```java
 class Solution {
     public TreeNode deleteNode(TreeNode root, int key) {
@@ -296,34 +324,59 @@ class Solution {
     }
 }
 ```
+递归法
 ```java
-// 解法2
 class Solution {
   public TreeNode deleteNode(TreeNode root, int key) {
-    if (root == null) return root;
-    if (root.val == key) {
-      if (root.left == null) {
-        return root.right;
-      } else if (root.right == null) {
-        return root.left;
-      } else {
-        TreeNode cur = root.right;
-        while (cur.left != null) {
-          cur = cur.left;
-        }
-        cur.left = root.left;
-        root = root.right;
-        return root;
+    if (root == null){
+      return null;
+    }
+    //寻找对应的对应的前面的节点，以及他的前一个节点
+    TreeNode cur = root;
+    TreeNode pre = null;
+    while (cur != null){
+      if (cur.val < key){
+        pre = cur;
+        cur = cur.right;
+      } else if (cur.val > key) {
+        pre = cur;
+        cur = cur.left;
+      }else {
+        break;
       }
     }
-    if (root.val > key) root.left = deleteNode(root.left, key);
-    if (root.val < key) root.right = deleteNode(root.right, key);
+    if (pre == null){
+      return deleteOneNode(cur);
+    }
+    if (pre.left !=null && pre.left.val == key){
+      pre.left = deleteOneNode(cur);
+    }
+    if (pre.right !=null && pre.right.val == key){
+      pre.right = deleteOneNode(cur);
+    }
     return root;
   }
+
+  public TreeNode deleteOneNode(TreeNode node){
+    if (node == null){
+      return null;
+    }
+    if (node.right == null){
+      return node.left;
+    }
+    TreeNode cur = node.right;
+    while (cur.left !=null){
+      cur = cur.left;
+    }
+    cur.left = node.left;
+    return node.right;
+  }
 }
 ```
 
+
 ### Python 
+
 递归法（版本一）
 ```python
 class Solution:
diff --git a/problems/0496.下一个更大元素I.md b/problems/0496.下一个更大元素I.md
index 31c3ce43..411a47df 100644
--- a/problems/0496.下一个更大元素I.md
+++ b/problems/0496.下一个更大元素I.md
@@ -387,6 +387,32 @@ function nextGreaterElement(nums1: number[], nums2: number[]): number[] {
 };
 ```
 
+Rust
+
+```rust
+impl Solution {
+    pub fn next_greater_element(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
+        let mut ans = vec![-1; nums1.len()];
+        use std::collections::HashMap;
+        let mut map = HashMap::new();
+        for (idx, &i) in nums1.iter().enumerate() {
+            map.insert(i, idx);
+        }
+        let mut stack = vec![];
+        for (idx, &i) in nums2.iter().enumerate() {
+            while !stack.is_empty() && nums2[*stack.last().unwrap()] < i {
+                let pos = stack.pop().unwrap();
+                if let Some(&jdx) = map.get(&nums2[pos]) {
+                    ans[jdx] = i;
+                }
+            }
+            stack.push(idx);
+        }
+        ans
+    }
+}
+```
+
 
 
 <p align="center">
diff --git a/problems/0739.每日温度.md b/problems/0739.每日温度.md
index 749dc972..d2da3737 100644
--- a/problems/0739.每日温度.md
+++ b/problems/0739.每日温度.md
@@ -455,7 +455,27 @@ function dailyTemperatures(temperatures: number[]): number[] {
 };
 ```
 
+Rust:
 
+```rust
+impl Solution {
+    /// 单调栈的本质是以空间换时间，记录之前已访问过的非递增子序列下标
+    pub fn daily_temperatures(temperatures: Vec<i32>) -> Vec<i32> {
+        let mut res = vec![0; temperatures.len()];
+        let mut stack = vec![];
+        for (idx, &value) in temperatures.iter().enumerate() {
+            while !stack.is_empty() && temperatures[*stack.last().unwrap()] < value {
+                // 弹出，并计算res中对应位置的值
+                let i = stack.pop().unwrap();
+                res[i] = (idx - i) as i32;
+            }
+            // 入栈
+            stack.push(idx)
+        }
+        res
+    }
+}
+```
 
 
 <p align="center">