feat: add solutions to lc problem: No.1026

solution/1000-1099/1026.Maximum Difference Between Node and Ancestor/README.md

@@ -50,7 +50,7 @@
 
 ### 方法一：DFS
 
-对于每个节点，求其与祖先节点的最大差值，我们只需要求出该节点与祖先节点最大值和最小值的差值，取所有差值的最大值即可。
+对于每个节点，求其与祖先节点的最大差值，我们只需要求出该节点与祖先节点最大值和最小值的差值。取所有节点与祖先节点差值的最大值即可。
 
 因此，我们设计一个函数 $dfs(root, mi, mx)$，表示当前搜索到的节点为 $root$，其祖先节点的最大值为 $mx$，最小值为 $mi$，函数内更新最大差值 $ans$。
 
@@ -75,7 +75,7 @@
 #         self.right = right
 class Solution:
     def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:
-        def dfs(root, mi, mx):
+        def dfs(root: Optional[TreeNode], mi: int, mx: int):
             if root is None:
                 return
             nonlocal ans
@@ -255,6 +255,42 @@ var maxAncestorDiff = function (root) {
 };
 ```
 
+```cs
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     public int val;
+ *     public TreeNode left;
+ *     public TreeNode right;
+ *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+public class Solution {
+    private int ans;
+
+    public int MaxAncestorDiff(TreeNode root) {
+        dfs(root, root.val, root.val);
+        return ans;
+    }
+
+    private void dfs(TreeNode root, int mi, int mx) {
+        if (root == null) {
+            return;
+        }
+        int x = Math.Max(Math.Abs(mi - root.val), Math.Abs(mx - root.val));
+        ans = Math.Max(ans, x);
+        mi = Math.Min(mi, root.val);
+        mx = Math.Max(mx, root.val);
+        dfs(root.left, mi, mx);
+        dfs(root.right, mi, mx);
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/1000-1099/1026.Maximum Difference Between Node and Ancestor/README_EN.md

@@ -40,7 +40,21 @@ Among all possible differences, the maximum value of 7 is obtained by |8 - 1| =
 
 ## Solutions
 
-### Solution 1
+### Solution 1: DFS
+
+For each node, to find the maximum difference with its ancestor nodes, we only need to find the difference between the maximum and minimum values of the ancestor nodes. The maximum difference among all nodes and their ancestor nodes is the answer.
+
+Therefore, we design a function $dfs(root, mi, mx)$, where the current node being searched is $root$, the maximum value of its ancestor nodes is $mx$, and the minimum value is $mi$. The function updates the maximum difference $ans$.
+
+The logic of the function $dfs(root, mi, mx)$ is as follows:
+
+-   If $root$ is null, return directly.
+-   Otherwise, we update $ans = max(ans, |mi - root.val|, |mx - root.val|)$.
+-   Then update $mi = min(mi, root.val)$, $mx = max(mx, root.val)$, and recursively search the left and right subtrees.
+
+In the main function, we call $dfs(root, root.val, root.val)$, and finally return $ans$.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the binary tree.
 
 <!-- tabs:start -->
 
@@ -53,7 +67,7 @@ Among all possible differences, the maximum value of 7 is obtained by |8 - 1| =
 #         self.right = right
 class Solution:
     def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:
-        def dfs(root, mi, mx):
+        def dfs(root: Optional[TreeNode], mi: int, mx: int):
             if root is None:
                 return
             nonlocal ans
@@ -233,6 +247,42 @@ var maxAncestorDiff = function (root) {
 };
 ```
 
+```cs
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     public int val;
+ *     public TreeNode left;
+ *     public TreeNode right;
+ *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+public class Solution {
+    private int ans;
+
+    public int MaxAncestorDiff(TreeNode root) {
+        dfs(root, root.val, root.val);
+        return ans;
+    }
+
+    private void dfs(TreeNode root, int mi, int mx) {
+        if (root == null) {
+            return;
+        }
+        int x = Math.Max(Math.Abs(mi - root.val), Math.Abs(mx - root.val));
+        ans = Math.Max(ans, x);
+        mi = Math.Min(mi, root.val);
+        mx = Math.Max(mx, root.val);
+        dfs(root.left, mi, mx);
+        dfs(root.right, mi, mx);
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/1000-1099/1026.Maximum Difference Between Node and Ancestor/Solution.cs

@@ -0,0 +1,33 @@
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     public int val;
+ *     public TreeNode left;
+ *     public TreeNode right;
+ *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+public class Solution {
+    private int ans;
+
+    public int MaxAncestorDiff(TreeNode root) {
+        dfs(root, root.val, root.val);
+        return ans;
+    }
+
+    private void dfs(TreeNode root, int mi, int mx) {
+        if (root == null) {
+            return;
+        }
+        int x = Math.Max(Math.Abs(mi - root.val), Math.Abs(mx - root.val));
+        ans = Math.Max(ans, x);
+        mi = Math.Min(mi, root.val);
+        mx = Math.Max(mx, root.val);
+        dfs(root.left, mi, mx);
+        dfs(root.right, mi, mx);
+    }
+}

solution/1000-1099/1026.Maximum Difference Between Node and Ancestor/Solution.py

@@ -6,7 +6,7 @@
 #         self.right = right
 class Solution:
     def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:
-        def dfs(root, mi, mx):
+        def dfs(root: Optional[TreeNode], mi: int, mx: int):
             if root is None:
                 return
             nonlocal ans