diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index 5f69f53d..1743243d 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -319,6 +319,36 @@ func levelOrder(_ root: TreeNode?) -> [[Int]] {
 }
 ```
 
+Rust:
+
+```rust
+pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
+    let mut ans = Vec::new();
+    let mut stack = Vec::new();
+    if root.is_none(){
+        return ans;
+    }
+    stack.push(root.unwrap());
+    while stack.is_empty()!= true{
+        let num = stack.len();
+        let mut level = Vec::new();
+        for _i in 0..num{
+            let tmp = stack.remove(0);
+            level.push(tmp.borrow_mut().val);
+            if tmp.borrow_mut().left.is_some(){
+                stack.push(tmp.borrow_mut().left.take().unwrap());
+            }
+            if tmp.borrow_mut().right.is_some(){
+                stack.push(tmp.borrow_mut().right.take().unwrap());
+            }
+        }
+        ans.push(level);
+    }
+    ans
+}
+```
+
+
 **此时我们就掌握了二叉树的层序遍历了，那么如下九道力扣上的题目，只需要修改模板的两三行代码（不能再多了），便可打倒！**
 
 
@@ -548,6 +578,35 @@ func levelOrderBottom(_ root: TreeNode?) -> [[Int]] {
 }
 ```
 
+Rust:
+
+```rust
+pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
+    let mut ans = Vec::new();
+    let mut stack = Vec::new();
+    if root.is_none(){
+        return ans;
+    }
+    stack.push(root.unwrap());
+    while stack.is_empty()!= true{
+        let num = stack.len();
+        let mut level = Vec::new();
+        for _i in 0..num{
+            let tmp = stack.remove(0);
+            level.push(tmp.borrow_mut().val);
+            if tmp.borrow_mut().left.is_some(){
+                stack.push(tmp.borrow_mut().left.take().unwrap());
+            }
+            if tmp.borrow_mut().right.is_some(){
+                stack.push(tmp.borrow_mut().right.take().unwrap());
+            }
+        }
+        ans.push(level);
+    }
+    ans
+}
+```
+
 # 199.二叉树的右视图
 
 [力扣题目链接](https://leetcode-cn.com/problems/binary-tree-right-side-view/)
diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md
index 2229a854..65a155fb 100644
--- a/problems/0104.二叉树的最大深度.md
+++ b/problems/0104.二叉树的最大深度.md
@@ -192,6 +192,33 @@ public:
 };
 ```
 
+rust:
+```rust
+impl Solution {
+    pub fn max_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
+        if root.is_none(){
+            return 0;
+        }
+        let mut max_depth: i32 = 0;
+        let mut stack = vec![root.unwrap()];
+        while !stack.is_empty() {
+            let num = stack.len();
+            for _i in 0..num{
+                let top = stack.remove(0);
+                if top.borrow_mut().left.is_some(){
+                    stack.push(top.borrow_mut().left.take().unwrap());
+                }
+                if top.borrow_mut().right.is_some(){
+                    stack.push(top.borrow_mut().right.take().unwrap());
+                }
+            }
+            max_depth+=1;
+        }
+        max_depth
+    }
+```
+
+
 那么我们可以顺便解决一下n叉树的最大深度问题
 
 # 559.n叉树的最大深度
diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md
index 224caa5e..b1331659 100644
--- a/problems/0111.二叉树的最小深度.md
+++ b/problems/0111.二叉树的最小深度.md
@@ -488,5 +488,69 @@ func minDepth(_ root: TreeNode?) -> Int {
 }
 ```
 
+rust:
+```rust
+impl Solution {
+    pub fn min_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
+        return Solution::bfs(root)
+    }
+
+    // 递归
+    pub fn dfs(node: Option<Rc<RefCell<TreeNode>>>) -> i32{
+        if node.is_none(){
+            return 0;
+        }
+        let parent = node.unwrap();
+        let left_child = parent.borrow_mut().left.take();
+        let right_child = parent.borrow_mut().right.take();
+        if left_child.is_none() && right_child.is_none(){
+            return 1;
+        }
+        let mut min_depth = i32::MAX;
+        if left_child.is_some(){
+            let left_depth = Solution::dfs(left_child);
+            if left_depth <= min_depth{
+                min_depth = left_depth
+            }
+        }
+        if right_child.is_some(){
+            let right_depth = Solution::dfs(right_child);
+            if right_depth <= min_depth{
+                min_depth = right_depth
+            }
+        }
+        min_depth + 1
+
+    }
+
+    // 迭代
+    pub fn bfs(node: Option<Rc<RefCell<TreeNode>>>) -> i32{
+        let mut min_depth = 0;
+        if node.is_none(){
+            return min_depth
+        }
+        let mut stack = vec![node.unwrap()];
+        while !stack.is_empty(){
+            min_depth += 1;
+            let num = stack.len();
+            for _i in 0..num{
+                let top = stack.remove(0);
+                let left_child = top.borrow_mut().left.take();
+                let right_child = top.borrow_mut().right.take();
+                if left_child.is_none() && right_child.is_none(){
+                    return min_depth;
+                }
+                if left_child.is_some(){
+                    stack.push(left_child.unwrap());
+                }
+                if right_child.is_some(){
+                    stack.push(right_child.unwrap());
+                }
+            }
+        }
+        min_depth
+    }
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>