diff --git a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/README.md b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/README.md
index 2e62530b63567..983c117496caa 100644
--- a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/README.md	
+++ b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/README.md	
@@ -59,7 +59,11 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：递归
+
+我们首先判断 $\textit{root}$ 是否为空，若为空则返回 0。否则我们初始化一个变量 $\textit{mx}$ 用来记录子节点的最大深度，然后遍历 $\textit{root}$ 的所有子节点，递归调用 $\text{maxDepth}$ 函数，更新 $\textit{mx}$ 的值。最后返回 $\textit{mx} + 1$ 即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为节点的数量。
 
 <!-- tabs:start -->
 
@@ -69,17 +73,20 @@ tags:
 """
 # Definition for a Node.
 class Node:
-    def __init__(self, val=None, children=None):
+    def __init__(self, val: Optional[int] = None, children: Optional[List['Node']] = None):
         self.val = val
         self.children = children
 """
 
 
 class Solution:
-    def maxDepth(self, root: 'Node') -> int:
+    def maxDepth(self, root: "Node") -> int:
         if root is None:
             return 0
-        return 1 + max([self.maxDepth(child) for child in root.children], default=0)
+        mx = 0
+        for child in root.children:
+            mx = max(mx, self.maxDepth(child))
+        return 1 + mx
 ```
 
 #### Java
@@ -109,11 +116,11 @@ class Solution {
         if (root == null) {
             return 0;
         }
-        int ans = 1;
+        int mx = 0;
         for (Node child : root.children) {
-            ans = Math.max(ans, 1 + maxDepth(child));
+            mx = Math.max(mx, maxDepth(child));
         }
-        return ans;
+        return 1 + mx;
     }
 }
 ```
@@ -144,10 +151,14 @@ public:
 class Solution {
 public:
     int maxDepth(Node* root) {
-        if (!root) return 0;
-        int ans = 1;
-        for (auto& child : root->children) ans = max(ans, 1 + maxDepth(child));
-        return ans;
+        if (!root) {
+            return 0;
+        }
+        int mx = 0;
+        for (Node* child : root->children) {
+            mx = max(mx, maxDepth(child));
+        }
+        return mx + 1;
     }
 };
 ```
@@ -167,11 +178,35 @@ func maxDepth(root *Node) int {
 	if root == nil {
 		return 0
 	}
-	ans := 1
+	mx := 0
 	for _, child := range root.Children {
-		ans = max(ans, 1+maxDepth(child))
+		mx = max(mx, maxDepth(child))
 	}
-	return ans
+	return 1 + mx
+}
+```
+
+#### TypeScript
+
+```ts
+/**
+ * Definition for _Node.
+ * class _Node {
+ *     val: number
+ *     children: _Node[]
+ *
+ *     constructor(val?: number, children?: _Node[]) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.children = (children===undefined ? [] : children)
+ *     }
+ * }
+ */
+
+function maxDepth(root: _Node | null): number {
+    if (!root) {
+        return 0;
+    }
+    return 1 + Math.max(...root.children.map(child => maxDepth(child)), 0);
 }
 ```
 
diff --git a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/README_EN.md b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/README_EN.md
index e22ba845c876f..00f282179e710 100644
--- a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/README_EN.md	
+++ b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/README_EN.md	
@@ -57,7 +57,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Recursion
+
+First, we check if $\textit{root}$ is null. If it is, we return 0. Otherwise, we initialize a variable $\textit{mx}$ to record the maximum depth of the child nodes, then traverse all the child nodes of $\textit{root}$, recursively call the $\text{maxDepth}$ function, and update the value of $\textit{mx}$. Finally, we return $\textit{mx} + 1$.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes.
 
 <!-- tabs:start -->
 
@@ -67,17 +71,20 @@ tags:
 """
 # Definition for a Node.
 class Node:
-    def __init__(self, val=None, children=None):
+    def __init__(self, val: Optional[int] = None, children: Optional[List['Node']] = None):
         self.val = val
         self.children = children
 """
 
 
 class Solution:
-    def maxDepth(self, root: 'Node') -> int:
+    def maxDepth(self, root: "Node") -> int:
         if root is None:
             return 0
-        return 1 + max([self.maxDepth(child) for child in root.children], default=0)
+        mx = 0
+        for child in root.children:
+            mx = max(mx, self.maxDepth(child))
+        return 1 + mx
 ```
 
 #### Java
@@ -107,11 +114,11 @@ class Solution {
         if (root == null) {
             return 0;
         }
-        int ans = 1;
+        int mx = 0;
         for (Node child : root.children) {
-            ans = Math.max(ans, 1 + maxDepth(child));
+            mx = Math.max(mx, maxDepth(child));
         }
-        return ans;
+        return 1 + mx;
     }
 }
 ```
@@ -142,10 +149,14 @@ public:
 class Solution {
 public:
     int maxDepth(Node* root) {
-        if (!root) return 0;
-        int ans = 1;
-        for (auto& child : root->children) ans = max(ans, 1 + maxDepth(child));
-        return ans;
+        if (!root) {
+            return 0;
+        }
+        int mx = 0;
+        for (Node* child : root->children) {
+            mx = max(mx, maxDepth(child));
+        }
+        return mx + 1;
     }
 };
 ```
@@ -165,11 +176,35 @@ func maxDepth(root *Node) int {
 	if root == nil {
 		return 0
 	}
-	ans := 1
+	mx := 0
 	for _, child := range root.Children {
-		ans = max(ans, 1+maxDepth(child))
+		mx = max(mx, maxDepth(child))
 	}
-	return ans
+	return 1 + mx
+}
+```
+
+#### TypeScript
+
+```ts
+/**
+ * Definition for _Node.
+ * class _Node {
+ *     val: number
+ *     children: _Node[]
+ *
+ *     constructor(val?: number, children?: _Node[]) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.children = (children===undefined ? [] : children)
+ *     }
+ * }
+ */
+
+function maxDepth(root: _Node | null): number {
+    if (!root) {
+        return 0;
+    }
+    return 1 + Math.max(...root.children.map(child => maxDepth(child)), 0);
 }
 ```
 
diff --git a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.cpp b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.cpp
index c2d959a1940d1..75e889f65d292 100644
--- a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.cpp	
+++ b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.cpp	
@@ -21,9 +21,13 @@ class Node {
 class Solution {
 public:
     int maxDepth(Node* root) {
-        if (!root) return 0;
-        int ans = 1;
-        for (auto& child : root->children) ans = max(ans, 1 + maxDepth(child));
-        return ans;
+        if (!root) {
+            return 0;
+        }
+        int mx = 0;
+        for (Node* child : root->children) {
+            mx = max(mx, maxDepth(child));
+        }
+        return mx + 1;
     }
-};
\ No newline at end of file
+};
diff --git a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.go b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.go
index 445451d86f048..8aaf065bb80da 100644
--- a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.go	
+++ b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.go	
@@ -10,9 +10,9 @@ func maxDepth(root *Node) int {
 	if root == nil {
 		return 0
 	}
-	ans := 1
+	mx := 0
 	for _, child := range root.Children {
-		ans = max(ans, 1+maxDepth(child))
+		mx = max(mx, maxDepth(child))
 	}
-	return ans
-}
\ No newline at end of file
+	return 1 + mx
+}
diff --git a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.java b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.java
index b3fbf183f5424..111d0b781b188 100644
--- a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.java	
+++ b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.java	
@@ -22,10 +22,10 @@ public int maxDepth(Node root) {
         if (root == null) {
             return 0;
         }
-        int ans = 1;
+        int mx = 0;
         for (Node child : root.children) {
-            ans = Math.max(ans, 1 + maxDepth(child));
+            mx = Math.max(mx, maxDepth(child));
         }
-        return ans;
+        return 1 + mx;
     }
-}
\ No newline at end of file
+}
diff --git a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.py b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.py
index a29f3ed859130..98c1895eaade8 100644
--- a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.py	
+++ b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.py	
@@ -1,14 +1,17 @@
 """
 # Definition for a Node.
 class Node:
-    def __init__(self, val=None, children=None):
+    def __init__(self, val: Optional[int] = None, children: Optional[List['Node']] = None):
         self.val = val
         self.children = children
 """
 
 
 class Solution:
-    def maxDepth(self, root: 'Node') -> int:
+    def maxDepth(self, root: "Node") -> int:
         if root is None:
             return 0
-        return 1 + max([self.maxDepth(child) for child in root.children], default=0)
+        mx = 0
+        for child in root.children:
+            mx = max(mx, self.maxDepth(child))
+        return 1 + mx
diff --git a/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.ts b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.ts
new file mode 100644
index 0000000000000..d68a4b482906e
--- /dev/null
+++ b/solution/0500-0599/0559.Maximum Depth of N-ary Tree/Solution.ts	
@@ -0,0 +1,19 @@
+/**
+ * Definition for _Node.
+ * class _Node {
+ *     val: number
+ *     children: _Node[]
+ *
+ *     constructor(val?: number, children?: _Node[]) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.children = (children===undefined ? [] : children)
+ *     }
+ * }
+ */
+
+function maxDepth(root: _Node | null): number {
+    if (!root) {
+        return 0;
+    }
+    return 1 + Math.max(...root.children.map(child => maxDepth(child)), 0);
+}
diff --git a/solution/0500-0599/0562.Longest Line of Consecutive One in Matrix/README.md b/solution/0500-0599/0562.Longest Line of Consecutive One in Matrix/README.md
index 361ad5a3c3226..598d48f20fd11 100644
--- a/solution/0500-0599/0562.Longest Line of Consecutive One in Matrix/README.md	
+++ b/solution/0500-0599/0562.Longest Line of Consecutive One in Matrix/README.md	
@@ -68,7 +68,7 @@ tags:
 
 遍历矩阵，当遇到 $1$ 时，更新 $f[i][j][k]$ 的值。对于每个位置 $(i, j)$，我们只需要更新其四个方向的值即可。然后更新答案。
 
-时间复杂度 $O(m\times n)$，空间复杂度 $O(m\times n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。
+时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。
 
 <!-- tabs:start -->
 
diff --git a/solution/0500-0599/0562.Longest Line of Consecutive One in Matrix/README_EN.md b/solution/0500-0599/0562.Longest Line of Consecutive One in Matrix/README_EN.md
index 1de17cddf10cb..3b04ffa448fa9 100644
--- a/solution/0500-0599/0562.Longest Line of Consecutive One in Matrix/README_EN.md	
+++ b/solution/0500-0599/0562.Longest Line of Consecutive One in Matrix/README_EN.md	
@@ -54,7 +54,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+We define $f[i][j][k]$ to represent the length of the longest consecutive $1$s ending at $(i, j)$ in direction $k$. The value range of $k$ is $0, 1, 2, 3$, representing horizontal, vertical, diagonal, and anti-diagonal directions, respectively.
+
+> We can also use four 2D arrays to represent the length of the longest consecutive $1$s in the four directions.
+
+We traverse the matrix, and when we encounter $1$, we update the value of $f[i][j][k]$. For each position $(i, j)$, we only need to update the values in its four directions. Then we update the answer.
+
+The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns in the matrix, respectively.
 
 <!-- tabs:start -->
 
diff --git a/solution/0500-0599/0563.Binary Tree Tilt/README.md b/solution/0500-0599/0563.Binary Tree Tilt/README.md
index dc1ff5a05ab75..025499f3fa99d 100644
--- a/solution/0500-0599/0563.Binary Tree Tilt/README.md	
+++ b/solution/0500-0599/0563.Binary Tree Tilt/README.md	
@@ -75,7 +75,13 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：递归
+
+我们设计一个函数 $\text{dfs}$，用来计算以当前节点为根节点的子树的节点之和。在 $\text{dfs}$ 函数中，我们首先判断当前节点是否为空，若为空则返回 0。然后递归调用 $\text{dfs}$ 函数计算左子树的节点之和 $l$ 和右子树的节点之和 $r$。接着计算当前节点的坡度，即 $|l - r|$，并将其加到答案中。最后返回当前节点的节点之和 $l + r + \textit{root.val}$。
+
+在主函数中，我们初始化答案为 0，然后调用 $\text{dfs}$ 函数计算整个树的坡度，并返回答案。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为节点的数量。
 
 <!-- tabs:start -->
 
@@ -89,19 +95,17 @@ tags:
 #         self.left = left
 #         self.right = right
 class Solution:
-    def findTilt(self, root: TreeNode) -> int:
-        ans = 0
-
-        def sum(root):
+    def findTilt(self, root: Optional[TreeNode]) -> int:
+        def dfs(root: Optional[TreeNode]) -> int:
             if root is None:
                 return 0
+            l, r = dfs(root.left), dfs(root.right)
             nonlocal ans
-            left = sum(root.left)
-            right = sum(root.right)
-            ans += abs(left - right)
-            return root.val + left + right
+            ans += abs(l - r)
+            return l + r + root.val
 
-        sum(root)
+        ans = 0
+        dfs(root)
         return ans
 ```
 
@@ -127,19 +131,17 @@ class Solution {
     private int ans;
 
     public int findTilt(TreeNode root) {
-        ans = 0;
-        sum(root);
+        dfs(root);
         return ans;
     }
 
-    private int sum(TreeNode root) {
+    private int dfs(TreeNode root) {
         if (root == null) {
             return 0;
         }
-        int left = sum(root.left);
-        int right = sum(root.right);
-        ans += Math.abs(left - right);
-        return root.val + left + right;
+        int l = dfs(root.left), r = dfs(root.right);
+        ans += Math.abs(l - r);
+        return l + r + root.val;
     }
 }
 ```
@@ -160,20 +162,19 @@ class Solution {
  */
 class Solution {
 public:
-    int ans;
-
     int findTilt(TreeNode* root) {
-        ans = 0;
-        sum(root);
+        int ans = 0;
+        auto dfs = [&](this auto&& dfs, TreeNode* root) -> int {
+            if (!root) {
+                return 0;
+            }
+            int l = dfs(root->left), r = dfs(root->right);
+            ans += abs(l - r);
+            return l + r + root->val;
+        };
+        dfs(root);
         return ans;
     }
-
-    int sum(TreeNode* root) {
-        if (!root) return 0;
-        int left = sum(root->left), right = sum(root->right);
-        ans += abs(left - right);
-        return root->val + left + right;
-    }
 };
 ```
 
@@ -188,28 +189,57 @@ public:
  *     Right *TreeNode
  * }
  */
-var ans int
-
-func findTilt(root *TreeNode) int {
-	ans = 0
-	sum(root)
-	return ans
-}
-
-func sum(root *TreeNode) int {
-	if root == nil {
-		return 0
+func findTilt(root *TreeNode) (ans int) {
+	var dfs func(*TreeNode) int
+	dfs = func(root *TreeNode) int {
+		if root == nil {
+			return 0
+		}
+		l, r := dfs(root.Left), dfs(root.Right)
+		ans += abs(l - r)
+		return l + r + root.Val
 	}
-	left, right := sum(root.Left), sum(root.Right)
-	ans += abs(left - right)
-	return root.Val + left + right
+	dfs(root)
+	return
 }
 
 func abs(x int) int {
-	if x > 0 {
-		return x
+	if x < 0 {
+		return -x
 	}
-	return -x
+	return x
+}
+```
+
+#### TypeScript
+
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function findTilt(root: TreeNode | null): number {
+    let ans: number = 0;
+    const dfs = (root: TreeNode | null): number => {
+        if (!root) {
+            return 0;
+        }
+        const [l, r] = [dfs(root.left), dfs(root.right)];
+        ans += Math.abs(l - r);
+        return l + r + root.val;
+    };
+    dfs(root);
+    return ans;
 }
 ```
 
diff --git a/solution/0500-0599/0563.Binary Tree Tilt/README_EN.md b/solution/0500-0599/0563.Binary Tree Tilt/README_EN.md
index ac323a07bd1f4..818f0d44432be 100644
--- a/solution/0500-0599/0563.Binary Tree Tilt/README_EN.md	
+++ b/solution/0500-0599/0563.Binary Tree Tilt/README_EN.md	
@@ -28,7 +28,7 @@ tags:
 <pre>
 <strong>Input:</strong> root = [1,2,3]
 <strong>Output:</strong> 1
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 Tilt of node 2 : |0-0| = 0 (no children)
 Tilt of node 3 : |0-0| = 0 (no children)
 Tilt of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3)
@@ -40,7 +40,7 @@ Sum of every tilt : 0 + 0 + 1 = 1
 <pre>
 <strong>Input:</strong> root = [4,2,9,3,5,null,7]
 <strong>Output:</strong> 15
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 Tilt of node 3 : |0-0| = 0 (no children)
 Tilt of node 5 : |0-0| = 0 (no children)
 Tilt of node 7 : |0-0| = 0 (no children)
@@ -71,7 +71,13 @@ Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Recursion
+
+We design a function $\text{dfs}$ to calculate the sum of nodes in the subtree rooted at the current node. In the $\text{dfs}$ function, we first check if the current node is null. If it is, we return 0. Then we recursively call the $\text{dfs}$ function to calculate the sum of nodes in the left subtree $l$ and the sum of nodes in the right subtree $r$. Next, we calculate the tilt of the current node, which is $|l - r|$, and add it to the answer. Finally, we return the sum of nodes of the current node, which is $l + r + \textit{root.val}$.
+
+In the main function, we initialize the answer to 0, then call the $\text{dfs}$ function to calculate the tilt of the entire tree and return the answer.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes.
 
 <!-- tabs:start -->
 
@@ -85,19 +91,17 @@ Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15
 #         self.left = left
 #         self.right = right
 class Solution:
-    def findTilt(self, root: TreeNode) -> int:
-        ans = 0
-
-        def sum(root):
+    def findTilt(self, root: Optional[TreeNode]) -> int:
+        def dfs(root: Optional[TreeNode]) -> int:
             if root is None:
                 return 0
+            l, r = dfs(root.left), dfs(root.right)
             nonlocal ans
-            left = sum(root.left)
-            right = sum(root.right)
-            ans += abs(left - right)
-            return root.val + left + right
+            ans += abs(l - r)
+            return l + r + root.val
 
-        sum(root)
+        ans = 0
+        dfs(root)
         return ans
 ```
 
@@ -123,19 +127,17 @@ class Solution {
     private int ans;
 
     public int findTilt(TreeNode root) {
-        ans = 0;
-        sum(root);
+        dfs(root);
         return ans;
     }
 
-    private int sum(TreeNode root) {
+    private int dfs(TreeNode root) {
         if (root == null) {
             return 0;
         }
-        int left = sum(root.left);
-        int right = sum(root.right);
-        ans += Math.abs(left - right);
-        return root.val + left + right;
+        int l = dfs(root.left), r = dfs(root.right);
+        ans += Math.abs(l - r);
+        return l + r + root.val;
     }
 }
 ```
@@ -156,20 +158,19 @@ class Solution {
  */
 class Solution {
 public:
-    int ans;
-
     int findTilt(TreeNode* root) {
-        ans = 0;
-        sum(root);
+        int ans = 0;
+        auto dfs = [&](this auto&& dfs, TreeNode* root) -> int {
+            if (!root) {
+                return 0;
+            }
+            int l = dfs(root->left), r = dfs(root->right);
+            ans += abs(l - r);
+            return l + r + root->val;
+        };
+        dfs(root);
         return ans;
     }
-
-    int sum(TreeNode* root) {
-        if (!root) return 0;
-        int left = sum(root->left), right = sum(root->right);
-        ans += abs(left - right);
-        return root->val + left + right;
-    }
 };
 ```
 
@@ -184,28 +185,57 @@ public:
  *     Right *TreeNode
  * }
  */
-var ans int
-
-func findTilt(root *TreeNode) int {
-	ans = 0
-	sum(root)
-	return ans
-}
-
-func sum(root *TreeNode) int {
-	if root == nil {
-		return 0
+func findTilt(root *TreeNode) (ans int) {
+	var dfs func(*TreeNode) int
+	dfs = func(root *TreeNode) int {
+		if root == nil {
+			return 0
+		}
+		l, r := dfs(root.Left), dfs(root.Right)
+		ans += abs(l - r)
+		return l + r + root.Val
 	}
-	left, right := sum(root.Left), sum(root.Right)
-	ans += abs(left - right)
-	return root.Val + left + right
+	dfs(root)
+	return
 }
 
 func abs(x int) int {
-	if x > 0 {
-		return x
+	if x < 0 {
+		return -x
 	}
-	return -x
+	return x
+}
+```
+
+#### TypeScript
+
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function findTilt(root: TreeNode | null): number {
+    let ans: number = 0;
+    const dfs = (root: TreeNode | null): number => {
+        if (!root) {
+            return 0;
+        }
+        const [l, r] = [dfs(root.left), dfs(root.right)];
+        ans += Math.abs(l - r);
+        return l + r + root.val;
+    };
+    dfs(root);
+    return ans;
 }
 ```
 
diff --git a/solution/0500-0599/0563.Binary Tree Tilt/Solution.cpp b/solution/0500-0599/0563.Binary Tree Tilt/Solution.cpp
index 1ca20e7cab9ee..6dccdaa121de6 100644
--- a/solution/0500-0599/0563.Binary Tree Tilt/Solution.cpp	
+++ b/solution/0500-0599/0563.Binary Tree Tilt/Solution.cpp	
@@ -11,18 +11,17 @@
  */
 class Solution {
 public:
-    int ans;
-
     int findTilt(TreeNode* root) {
-        ans = 0;
-        sum(root);
+        int ans = 0;
+        auto dfs = [&](this auto&& dfs, TreeNode* root) -> int {
+            if (!root) {
+                return 0;
+            }
+            int l = dfs(root->left), r = dfs(root->right);
+            ans += abs(l - r);
+            return l + r + root->val;
+        };
+        dfs(root);
         return ans;
     }
-
-    int sum(TreeNode* root) {
-        if (!root) return 0;
-        int left = sum(root->left), right = sum(root->right);
-        ans += abs(left - right);
-        return root->val + left + right;
-    }
-};
\ No newline at end of file
+};
diff --git a/solution/0500-0599/0563.Binary Tree Tilt/Solution.go b/solution/0500-0599/0563.Binary Tree Tilt/Solution.go
index 1fc9c98235902..c3956d8d6a7ae 100644
--- a/solution/0500-0599/0563.Binary Tree Tilt/Solution.go	
+++ b/solution/0500-0599/0563.Binary Tree Tilt/Solution.go	
@@ -6,26 +6,23 @@
  *     Right *TreeNode
  * }
  */
-var ans int
-
-func findTilt(root *TreeNode) int {
-	ans = 0
-	sum(root)
-	return ans
-}
-
-func sum(root *TreeNode) int {
-	if root == nil {
-		return 0
+func findTilt(root *TreeNode) (ans int) {
+	var dfs func(*TreeNode) int
+	dfs = func(root *TreeNode) int {
+		if root == nil {
+			return 0
+		}
+		l, r := dfs(root.Left), dfs(root.Right)
+		ans += abs(l - r)
+		return l + r + root.Val
 	}
-	left, right := sum(root.Left), sum(root.Right)
-	ans += abs(left - right)
-	return root.Val + left + right
+	dfs(root)
+	return
 }
 
 func abs(x int) int {
-	if x > 0 {
-		return x
+	if x < 0 {
+		return -x
 	}
-	return -x
-}
\ No newline at end of file
+	return x
+}
diff --git a/solution/0500-0599/0563.Binary Tree Tilt/Solution.java b/solution/0500-0599/0563.Binary Tree Tilt/Solution.java
index f03febf02b52b..1db5d9c38bdfc 100644
--- a/solution/0500-0599/0563.Binary Tree Tilt/Solution.java	
+++ b/solution/0500-0599/0563.Binary Tree Tilt/Solution.java	
@@ -17,18 +17,16 @@ class Solution {
     private int ans;
 
     public int findTilt(TreeNode root) {
-        ans = 0;
-        sum(root);
+        dfs(root);
         return ans;
     }
 
-    private int sum(TreeNode root) {
+    private int dfs(TreeNode root) {
         if (root == null) {
             return 0;
         }
-        int left = sum(root.left);
-        int right = sum(root.right);
-        ans += Math.abs(left - right);
-        return root.val + left + right;
+        int l = dfs(root.left), r = dfs(root.right);
+        ans += Math.abs(l - r);
+        return l + r + root.val;
     }
-}
\ No newline at end of file
+}
diff --git a/solution/0500-0599/0563.Binary Tree Tilt/Solution.py b/solution/0500-0599/0563.Binary Tree Tilt/Solution.py
index 96fb38e618e63..0f799a54101c5 100644
--- a/solution/0500-0599/0563.Binary Tree Tilt/Solution.py	
+++ b/solution/0500-0599/0563.Binary Tree Tilt/Solution.py	
@@ -5,17 +5,15 @@
 #         self.left = left
 #         self.right = right
 class Solution:
-    def findTilt(self, root: TreeNode) -> int:
-        ans = 0
-
-        def sum(root):
+    def findTilt(self, root: Optional[TreeNode]) -> int:
+        def dfs(root: Optional[TreeNode]) -> int:
             if root is None:
                 return 0
+            l, r = dfs(root.left), dfs(root.right)
             nonlocal ans
-            left = sum(root.left)
-            right = sum(root.right)
-            ans += abs(left - right)
-            return root.val + left + right
+            ans += abs(l - r)
+            return l + r + root.val
 
-        sum(root)
+        ans = 0
+        dfs(root)
         return ans
diff --git a/solution/0500-0599/0563.Binary Tree Tilt/Solution.ts b/solution/0500-0599/0563.Binary Tree Tilt/Solution.ts
new file mode 100644
index 0000000000000..cc5f70a65886b
--- /dev/null
+++ b/solution/0500-0599/0563.Binary Tree Tilt/Solution.ts	
@@ -0,0 +1,27 @@
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function findTilt(root: TreeNode | null): number {
+    let ans: number = 0;
+    const dfs = (root: TreeNode | null): number => {
+        if (!root) {
+            return 0;
+        }
+        const [l, r] = [dfs(root.left), dfs(root.right)];
+        ans += Math.abs(l - r);
+        return l + r + root.val;
+    };
+    dfs(root);
+    return ans;
+}
diff --git a/solution/0500-0599/0566.Reshape the Matrix/README_EN.md b/solution/0500-0599/0566.Reshape the Matrix/README_EN.md
index fa4bc220d8cd8..a8b4f4f528e77 100644
--- a/solution/0500-0599/0566.Reshape the Matrix/README_EN.md	
+++ b/solution/0500-0599/0566.Reshape the Matrix/README_EN.md	
@@ -58,7 +58,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+First, we get the number of rows and columns of the original matrix, denoted as $m$ and $n$ respectively. If $m \times n \neq r \times c$, then the matrix cannot be reshaped, and we return the original matrix directly.
+
+Otherwise, we create a new matrix with $r$ rows and $c$ columns. Starting from the first element of the original matrix, we traverse all elements in row-major order and place the traversed elements into the new matrix in order.
+
+After traversing all elements of the original matrix, we get the answer.
+
+The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the original matrix, respectively. Ignoring the space consumption of the answer, the space complexity is $O(1)$.
 
 <!-- tabs:start -->