[pull] master from youngyangyang04:master

problems/0101.对称二叉树.md

@@ -897,6 +897,53 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+// 递归
+public bool IsSymmetric(TreeNode root)
+{
+    if (root == null) return true;
+    return Compare(root.left, root.right);
+}
+public bool Compare(TreeNode left, TreeNode right)
+{
+    if(left == null && right != null) return false;
+    else if(left != null && right == null ) return false;
+    else if(left == null && right == null) return true;
+    else if(left.val != right.val) return false;
+
+    var outside = Compare(left.left, right.right);
+    var inside = Compare(left.right, right.left);
+
+    return outside&&inside;
+}
+```
+``` C#
+// 迭代法
+public bool IsSymmetric(TreeNode root)
+{
+    if (root == null) return true;
+    var st = new Stack<TreeNode>();
+    st.Push(root.left);
+    st.Push(root.right);
+    while (st.Count != 0)
+    {
+        var left = st.Pop();
+        var right = st.Pop();
+        if (left == null && right == null)
+            continue;
+
+        if ((left == null || right == null || (left.val != right.val)))
+            return false;
+
+        st.Push(left.left);
+        st.Push(right.right);
+        st.Push(left.right);
+        st.Push(right.left);
+    }
+    return true;
+}
+```
 
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problems/0102.二叉树的层序遍历.md

@@ -462,6 +462,32 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+public IList<IList<int>> LevelOrder(TreeNode root)
+{
+    var res = new List<IList<int>>();
+    var que = new Queue<TreeNode>();
+    if (root == null) return res;
+    que.Enqueue(root);
+    while (que.Count != 0)
+    {
+        var size = que.Count;
+        var vec = new List<int>();
+        for (int i = 0; i < size; i++)
+        {
+            var cur = que.Dequeue();
+            vec.Add(cur.val);
+            if (cur.left != null) que.Enqueue(cur.left);
+            if (cur.right != null) que.Enqueue(cur.right);
+        }
+        res.Add(vec);
+
+
+    }
+    return res;
+}
+```
 
 
 **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！**
@@ -798,6 +824,31 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+public IList<IList<int>> LevelOrderBottom(TreeNode root)
+{
+    var res = new List<IList<int>>();
+    var que = new Queue<TreeNode>();
+    if (root == null) return res;
+    que.Enqueue(root);
+    while (que.Count != 0)
+    {
+        var size = que.Count;
+        var vec = new List<int>();
+        for (int i = 0; i < size; i++)
+        {
+            var cur = que.Dequeue();
+            vec.Add(cur.val);
+            if (cur.left != null) que.Enqueue(cur.left);
+            if (cur.right != null) que.Enqueue(cur.right);
+        }
+        res.Add(vec);
+    }
+    res.Reverse();
+    return res;
+}
+```
 
 ## 199.二叉树的右视图
 

problems/0104.二叉树的最大深度.md

@@ -1022,6 +1022,61 @@ impl Solution {
         max_depth
     }
 ```
+### C#
+```C#
+// 递归法
+public int MaxDepth(TreeNode root) {
+    if(root == null) return 0;
+
+    int leftDepth = MaxDepth(root.left);
+    int rightDepth = MaxDepth(root.right);
+
+    return 1 + Math.Max(leftDepth, rightDepth);
+}
+```
+```C#
+// 前序遍历
+int result = 0;
+public int MaxDepth(TreeNode root)
+{
+    if (root == null) return result;
+    GetDepth(root, 1);
+    return result;
+}
+public void GetDepth(TreeNode root, int depth)
+{
+    result = depth > result ? depth : result;
+    if (root.left == null && root.right == null) return;
+
+    if (root.left != null)
+        GetDepth(root.left, depth + 1);
+    if (root.right != null)
+        GetDepth(root.right, depth + 1);
+    return;
+}
+```
+```C#
+// 迭代法
+public int MaxDepth(TreeNode root)
+{
+    int depth = 0;
+    Queue<TreeNode> que = new();
+    if (root == null) return depth;
+    que.Enqueue(root);
+    while (que.Count != 0)
+    {
+        int size = que.Count;
+        depth++;
+        for (int i = 0; i < size; i++)
+        {
+            var node = que.Dequeue();
+            if (node.left != null) que.Enqueue(node.left);
+            if (node.right != null) que.Enqueue(node.right);
+        }
+    }
+    return depth;
+}
+```
 
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problems/0226.翻转二叉树.md

@@ -1018,25 +1018,13 @@ public class Solution {
 ```csharp
 //迭代
 public class Solution {
-    public TreeNode InvertTree(TreeNode root) {
-          if (root == null) return null;
-        Stack<TreeNode> stack=new Stack<TreeNode>();
-        stack.Push(root);
-        while(stack.Count>0)
-        {
-            TreeNode node = stack.Pop();   
-            swap(node);
-            if(node.right!=null) stack.Push(node.right);   
-            if(node.left!=null) stack.Push(node.left);     
-        }
-        return root;
-    }
-
-    public void swap(TreeNode node) {
-        TreeNode temp = node.left;
-        node.left = node.right;
-        node.right = temp;
-    }
+public TreeNode InvertTree(TreeNode root) {
+    if(root == null) return root;
+    (root.left,root.right) = (root.right, root.left);
+    InvertTree(root.left);
+    InvertTree(root.right);
+    return root;
+}
 }
 ```
 

problems/0344.反转字符串.md

@@ -250,6 +250,20 @@ class Solution:
         s[:] = [s[i] for i in range(len(s) - 1, -1, -1)]
 
 ```
+
+（版本七） 使用reverse()
+
+```python
+class Solution:
+    def reverseString(self, s: List[str]) -> None:
+        """
+        Do not return anything, modify s in-place instead.
+        """
+        # 原地反转,无返回值
+        s.reverse()
+
+```
+
 ### Go：
 
 ```Go

problems/0383.赎金信.md

@@ -214,6 +214,19 @@ class Solution:
         return all(ransomNote.count(c) <= magazine.count(c) for c in set(ransomNote))
 ```
 
+(版本六）使用count(简单易懂)
+
+```python3
+class Solution:
+    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
+        for char in ransomNote:
+            if char in magazine and ransomNote.count(char) <= magazine.count(char):
+                continue
+            else:
+                return False
+        return True
+```
+
 ### Go：
 
 ```go

problems/二叉树的统一迭代法.md

@@ -741,6 +741,99 @@ impl Solution{
     }
 }
 ```
+### C#
+```C#
+// 前序遍历
+public IList<int> PreorderTraversal(TreeNode root)
+{
+    var res = new List<int>();
+    var st = new Stack<TreeNode>();
+    if (root == null) return res;
+    st.Push(root);
+    while (st.Count != 0)
+    {
+        var node = st.Peek();
+        if (node == null)
+        {
+            st.Pop();
+            node = st.Peek();
+            st.Pop();
+            res.Add(node.val);
+        }
+        else
+        {
+            st.Pop();
+            if (node.right != null) st.Push(node.right);
+            if (node.left != null) st.Push(node.left);
+            st.Push(node);
+            st.Push(null);
+        }
+    }
+    return res;
+}
+```
+```C#
+// 中序遍历
+public IList<int> InorderTraversal(TreeNode root)
+{
+    var res = new List<int>();
+    var st = new Stack<TreeNode>();
+    if (root == null) return res;
+    st.Push(root);
+    while (st.Count != 0)
+    {
+        var node = st.Peek();
+        if (node == null)
+        {
+            st.Pop();
+            node = st.Peek();
+            st.Pop();
+            res.Add(node.val);
+        }
+        else
+        {
+            st.Pop();
+            if (node.right != null) st.Push(node.right);
+            st.Push(node);
+            st.Push(null);
+            if (node.left != null) st.Push(node.left);
+        }
+    }
+    return res;
+}
+```
+
+```C#
+// 后序遍历
+public IList<int> PostorderTraversal(TreeNode root)
+{
+    var res = new List<int>();
+    var st = new Stack<TreeNode>();
+    if (root == null) return res;
+    st.Push(root);
+    while (st.Count != 0)
+    {
+        var node = st.Peek();
+        if (node == null)
+        {
+            st.Pop();
+            node = st.Peek();
+            st.Pop();
+            res.Add(node.val);
+        }
+        else
+        {
+            st.Pop();
+            if (node.left != null) st.Push(node.left);
+            if (node.right != null) st.Push(node.right);
+            st.Push(node);
+            st.Push(null);
+        }
+    }
+    res.Reverse(0, res.Count);
+    return res;
+}
+```
 
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