Merge branch 'master' of github.com:youngyangyang04/leetcode-master

README.md

@@ -397,9 +397,9 @@
 * [图论：417.太平洋大西洋水流问题](./problems/0417.太平洋大西洋水流问题.md)
 * [图论：827.最大人工岛](./problems/0827.最大人工岛.md)
 * [图论：127. 单词接龙](./problems/0127.单词接龙.md)
-* [图论：841.钥匙和房间](./problems/841.钥匙和房间)
+* [图论：841.钥匙和房间](./problems/0841.钥匙和房间.md)
 * [图论：463. 岛屿的周长](./problems/0463.岛屿的周长.md)
-* [图论：并查集理论基础](./problems/)
+* [图论：并查集理论基础](./problems/图论并查集理论基础.md)
 * [图论：1971. 寻找图中是否存在路径](./problems/1971.寻找图中是否存在路径.md)
 * [图论：684.冗余连接](./problems/0684.冗余连接.md)
 * [图论：685.冗余连接II](./problems/0685.冗余连接II.md)

problems/0070.爬楼梯完全背包版本.md

@@ -100,8 +100,8 @@ int main() {
     while (cin >> n >> m) {
         vector<int> dp(n + 1, 0);
         dp[0] = 1;
-        for (int i = 1; i <= n; i++) { // 遍历物品
+        for (int i = 1; i <= n; i++) { // 遍历背包
-            for (int j = 1; j <= m; j++) { // 遍历背包
+            for (int j = 1; j <= m; j++) { // 遍历物品
                 if (i - j >= 0) dp[i] += dp[i - j];
             }
         }

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;
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

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

@@ -738,26 +738,36 @@ function maxDepth(root: TreeNode | null): number {
 
 ```typescript
 // 后续遍历（自下而上）
-function maxDepth(root: TreeNode | null): number {
+function maxDepth(root: Node | null): number {
-    if (root === null) return 0;
+    if (root === null) return 0
-    return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
+    let depth = 0
-};
+    for (let i = 0; i < root.children.length; i++) {
+        depth = Math.max(depth, maxDepth(root.children[i]))
+    }
+    return depth + 1
+}
 
 // 前序遍历(自上而下)
-function maxDepth(root: TreeNode | null): number {
+function maxDepth(root: Node | null): number {
-    function recur(node: TreeNode | null, count: number) {
+    if (root === null) return 0
-        if (node === null) {
+
-            resMax = resMax > count ? resMax : count;
+    let depth: number = 0
-            return;
+    const queue: Array<Node | null> = []
+    queue.push(root)
+
+    while (queue.length > 0) {
+        let len = queue.length
+        depth++
+        for (let i = 0; i < len; i++) {
+            // 当前层遍历
+            let curNode: Node | null = queue.shift()!
+                for (let j = 0; j < curNode.children.length; j++) {
+                    if (curNode.children[j]) queue.push(curNode.children[j])
                 }
-        recur(node.left, count + 1);
-        recur(node.right, count + 1);
         }
-    let resMax: number = 0;
+    }
-    let count: number = 0;
+    return depth
-    recur(root, count);
+}
-    return resMax;
-};
 
 
 ```
@@ -1022,6 +1032,61 @@ 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;
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/0110.平衡二叉树.md

@@ -908,6 +908,31 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+public bool IsBalanced(TreeNode root)
+{
+    return GetHeight(root) == -1 ? false : true;
+}
+public int GetHeight(TreeNode root)
+{
+    if (root == null) return 0;
+    int left = GetHeight(root.left);
+    if (left == -1) return -1;
+    int right = GetHeight(root.right);
+    if (right == -1) return -1;
+    int res;
+    if (Math.Abs(left - right) > 1)
+    {
+        res = -1;
+    }
+    else
+    {
+        res = 1 + Math.Max(left, right);
+    }
+    return res;
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/0111.二叉树的最小深度.md

@@ -708,6 +708,49 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+// 递归
+public int MinDepth(TreeNode root)
+{
+    if (root == null) return 0;
+    int left = MinDepth(root.left);
+    int right = MinDepth(root.right);
+    if (root.left == null && root.right != null)
+        return 1+right;
+    else if(root.left!=null && root.right == null)
+        return 1+left;
+        
+    int res = 1 + Math.Min(left, right);
+    return res;
+}
+```
+```C#
+// 迭代
+public int MinDepth(TreeNode root)
+{
+    if (root == null) return 0;
+    int depth = 0;
+    var que = new Queue<TreeNode>();
+    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);
+            if (node.left == null && node.right == null)
+                return depth;
+        }
+    }
+    return depth;
+}     
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/0112.路径总和.md

@@ -1511,6 +1511,17 @@ impl Solution {
     }
 }
 
+```
+### C#
+```C#
+// 0112.路径总和
+// 递归 
+public bool HasPathSum(TreeNode root, int targetSum)
+{
+    if (root == null) return false;
+    if (root.left == null && root.right == null && targetSum == root.val) return true;
+    return HasPathSum(root.left, targetSum - root.val) || HasPathSum(root.right, targetSum - root.val);
+}
 ```
 
 <p align="center">

problems/0222.完全二叉树的节点个数.md

@@ -867,6 +867,31 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+// 递归
+public int CountNodes(TreeNode root)
+{
+    if (root == null) return 0;
+    var left = root.left;
+    var right = root.right;
+    int leftDepth = 0, rightDepth = 0;
+    while (left != null)
+    {
+        left = left.left;
+        leftDepth++;
+    }
+    while (right != null)
+    {
+        right = right.right;
+        rightDepth++;
+    }
+    if (leftDepth == rightDepth)
+        return (int)Math.Pow(2, leftDepth+1) - 1;
+    return CountNodes(root.left) + CountNodes(root.right) + 1;
+
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/0226.翻转二叉树.md

@@ -1019,24 +1019,12 @@ public class Solution {
 //迭代
 public class Solution {
 public TreeNode InvertTree(TreeNode root) {
-          if (root == null) return null;
+    if(root == null) return root;
-        Stack<TreeNode> stack=new Stack<TreeNode>();
+    (root.left,root.right) = (root.right, root.left);
-        stack.Push(root);
+    InvertTree(root.left);
-        while(stack.Count>0)
+    InvertTree(root.right);
-        {
-            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;
-    }
 }
 ```
 

problems/0257.二叉树的所有路径.md

@@ -900,6 +900,43 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+public IList<string> BinaryTreePaths(TreeNode root)
+{
+    List<int> path = new();
+    List<string> res = new();
+    if (root == null) return res;
+    Traversal(root, path, res);
+    return res;
+}
+public void Traversal(TreeNode node, List<int> path, List<string> res)
+{
+    path.Add(node.val);
+    if (node.left == null && node.right == null)
+    {
+        string sPath = "";
+        for (int i = 0; i < path.Count - 1; i++)
+        {
+            sPath += path[i].ToString();
+            sPath += "->";
+        }
+        sPath += path[path.Count - 1].ToString();
+        res.Add(sPath);
+        return;
+    }
+    if (node.left != null)
+    {
+        Traversal(node.left, path, res);
+        path.RemoveAt(path.Count-1);
+    }
+    if (node.right != null)
+    {
+        Traversal(node.right, path, res);
+        path.RemoveAt(path.Count-1);
+    }
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

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/0404.左叶子之和.md

@@ -651,6 +651,23 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+// 递归
+public int SumOfLeftLeaves(TreeNode root)
+{
+    if (root == null) return 0;
+
+    int leftValue = SumOfLeftLeaves(root.left);
+    if (root.left != null && root.left.left == null && root.left.right == null)
+    {
+        leftValue += root.left.val;
+    }
+    int rightValue = SumOfLeftLeaves(root.right);
+    return leftValue + rightValue;
+
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/0513.找树左下角的值.md

@@ -684,6 +684,38 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+//递归
+int maxDepth = -1;
+int res = 0;
+public int FindBottomLeftValue(TreeNode root)
+{
+    Traversal(root, 0);
+    return res;
+}
+public void Traversal(TreeNode root, int depth)
+{
+    if (root.left == null && root.right == null)
+    {
+        if (depth > maxDepth)
+        {
+            maxDepth = depth;
+            res = root.val;
+        }
+        return;
+    }
+    if (root.left != null)
+    {
+        Traversal(root.left, depth + 1);
+    }
+    if (root.right != null)
+    {
+        Traversal(root.right, depth + 1);
+    }
+    return;
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/kama54.替换数字.md

@@ -46,7 +46,7 @@
 
 从前向后填充就是O(n^2)的算法了，因为每次添加元素都要将添加元素之后的所有元素整体向后移动。
 
-**其实很多数组填充类的问题，其做饭都是先预先给数组扩容带填充后的大小，然后在从后向前进行操作。**
+**其实很多数组填充类的问题，其做法都是先预先给数组扩容带填充后的大小，然后在从后向前进行操作。**
 
 这么做有两个好处：
 
@@ -162,6 +162,27 @@ class Main {
 ```
 
 ### Go：
+````go
+package main
+
+import "fmt"
+
+func main(){
+    var strByte []byte
+    
+    fmt.Scanln(&strByte)
+    
+    for i := 0; i < len(strByte); i++{
+        if strByte[i] <= '9' && strByte[i] >= '0' {
+            inserElement := []byte{'n','u','m','b','e','r'}
+            strByte = append(strByte[:i], append(inserElement, strByte[i+1:]...)...)
+            i = i + len(inserElement) -1
+        }
+    }
+    
+    fmt.Printf(string(strByte))
+}
+````
 
 
 

problems/kama55.右旋字符串.md

@@ -214,6 +214,34 @@ public class Main {
 
 
 ### Go：
+```go
+package main
+import "fmt"
+
+func reverse (strByte []byte, l, r int){
+    for l < r {
+        strByte[l], strByte[r] = strByte[r], strByte[l]
+        l++
+        r--
+    }
+}
+
+
+func main(){
+    var str string
+    var target int
+    
+    fmt.Scanln(&target)
+    fmt.Scanln(&str)
+    strByte := []byte(str)
+    
+    reverse(strByte, 0, len(strByte) - 1)
+    reverse(strByte, 0, target - 1)
+    reverse(strByte, target, len(strByte) - 1)
+    
+    fmt.Printf(string(strByte))
+}
+```
 
 
 ### JavaScript：

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;
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/二叉树的迭代遍历.md

@@ -695,6 +695,67 @@ impl Solution {
     }
 }
 ```
+### C#
+```C#
+// 前序遍历
+public IList<int> PreorderTraversal(TreeNode root)
+{
+    var st = new Stack<TreeNode>();
+    var res = new List<int>();
+    if (root == null) return res;
+    st.Push(root);
+    while (st.Count != 0)
+    {
+        var node = st.Pop();
+        res.Add(node.val);
+        if (node.right != null)
+            st.Push(node.right);
+        if (node.left != null)
+            st.Push(node.left);
+    }
+    return res;
+}  
+
+// 中序遍历
+public IList<int> InorderTraversal(TreeNode root)
+{
+    var st = new Stack<TreeNode>();
+    var res = new List<int>();
+    var cur = root;
+    while (st.Count != 0 || cur != null)
+    {
+        if (cur != null)
+        {
+            st.Push(cur);
+            cur = cur.left;
+        }
+        else
+        {
+            cur = st.Pop();
+            res.Add(cur.val);
+            cur = cur.right;
+        }
+    }
+    return res;
+}
+// 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 cur = st.Pop();
+        res.Add(cur.val);
+        if (cur.left != null) st.Push(cur.left);
+        if (cur.right != null) st.Push(cur.right);
+    }
+    res.Reverse(0, res.Count());
+    return res;
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/二叉树的递归遍历.md

@@ -565,6 +565,60 @@ impl Solution {
 }
 ```
 
+### C#
+```C#
+// 前序遍历
+public IList<int> PreorderTraversal(TreeNode root)
+{
+    var res = new List<int>();
+    if (root == null) return res;
+    Traversal(root, res);
+    return res;
+
+}
+public void Traversal(TreeNode cur, IList<int> res)
+{
+    if (cur == null) return;
+    res.Add(cur.val);
+    Traversal(cur.left, res);
+    Traversal(cur.right, res);
+}
+```
+```C#
+// 中序遍历
+public IList<int> InorderTraversal(TreeNode root)
+{
+    var res = new List<int>();
+    if (root == null) return res;
+    Traversal(root, res);
+    return res;
+}
+public void Traversal(TreeNode cur, IList<int> res)
+{
+    if (cur == null) return;
+    Traversal(cur.left, res);
+    res.Add(cur.val);
+    Traversal(cur.right, res);
+}
+```
+```C#
+// 后序遍历
+public IList<int> PostorderTraversal(TreeNode root)
+{
+    var res = new List<int>();
+    if (root == null) return res;
+    Traversal(root, res);
+    return res;
+}
+public void Traversal(TreeNode cur, IList<int> res)
+{
+    if (cur == null) return;
+    Traversal(cur.left, res);
+    Traversal(cur.right, res);
+    res.Add(cur.val);
+}
+```
+
 <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"/>