feat: add solutions to lc problem: No.0098

No.0098.Validate Binary Search Tree

solution/0000-0099/0098.Validate Binary Search Tree/README.md

@@ -46,7 +46,13 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-中序遍历，若是一个有效的二叉搜索树，那么遍历到的序列应该是单调递增的。所以只要比较判断遍历到的当前数是否 `>=` 上一个数即可。
+**方法一：递归**
+
+中序遍历，若是一个有效的二叉搜索树，那么遍历到的序列应该是单调递增的。所以只要比较判断遍历到的当前数是否大于上一个数即可。
+
+或者考虑以 `root` 为根的子树，所有节点的值是否都在合法范围内，递归判断即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是树中节点的数量。
 
 <!-- tabs:start -->
 
@@ -62,7 +68,7 @@
 #         self.left = left
 #         self.right = right
 class Solution:
-    def isValidBST(self, root: TreeNode) -> bool:
+    def isValidBST(self, root: Optional[TreeNode]) -> bool:
         def dfs(root):
             nonlocal prev
             if root is None:
@@ -80,6 +86,25 @@ class Solution:
         return dfs(root)
 ```
 
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def isValidBST(self, root: Optional[TreeNode]) -> bool:
+        def dfs(root, l, r):
+            if root is None:
+                return True
+            if root.val <= l or root.val >= r:
+                return False
+            return dfs(root.left, l, root.val) and dfs(root.right, root.val, r)
+
+        return dfs(root, -inf, inf)
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -127,6 +152,39 @@ class Solution {
 }
 ```
 
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    public boolean isValidBST(TreeNode root) {
+        return dfs(root, Long.MIN_VALUE, Long.MAX_VALUE);
+    }
+
+    private boolean dfs(TreeNode root, long l, long r) {
+        if (root == null) {
+            return true;
+        }
+        if (root.val <= l || root.val >= r) {
+            return false;
+        }
+        return dfs(root.left, l, root.val) && dfs(root.right, root.val, r);
+    }
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -161,6 +219,32 @@ public:
 };
 ```
 
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    bool isValidBST(TreeNode* root) {
+        return dfs(root, LONG_MIN, LONG_MAX);
+    }
+
+    bool dfs(TreeNode* root, long long l, long long r) {
+        if (!root) return true;
+        if (root->val <= l || root->val >= r) return false;
+        return dfs(root->left, l, root->val) && dfs(root->right, root->val, r);
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -197,6 +281,31 @@ func isValidBST(root *TreeNode) bool {
 }
 ```
 
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func isValidBST(root *TreeNode) bool {
+	return dfs(root, math.MinInt64, math.MaxInt64)
+}
+
+func dfs(root *TreeNode, l, r int64) bool {
+	if root == nil {
+		return true
+	}
+	v := int64(root.Val)
+	if v <= l || v >= r {
+		return false
+	}
+	return dfs(root.Left, l, v) && dfs(root.Right, v, r)
+}
+```
+
 ### **JavaScript**
 
 ```js
@@ -236,6 +345,33 @@ var isValidBST = function (root) {
 };
 ```
 
+```js
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {boolean}
+ */
+var isValidBST = function (root) {
+    function dfs(root, l, r) {
+        if (!root) {
+            return true;
+        }
+        if (root.val <= l || root.val >= r) {
+            return false;
+        }
+        return dfs(root.left, l, root.val) && dfs(root.right, root.val, r);
+    }
+    return dfs(root, -Infinity, Infinity);
+};
+```
+
 ### **C#**
 
 ```cs
@@ -283,6 +419,37 @@ public class Solution {
 }
 ```
 
+```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 {
+    public bool IsValidBST(TreeNode root) {
+        return dfs(root, long.MinValue, long.MaxValue);
+    }
+
+    public bool dfs(TreeNode root, long l, long r) {
+        if (root == null) {
+            return true;
+        }
+        if (root.val <= l || root.val >= r) {
+            return false;
+        }
+        return dfs(root.left, l, root.val) && dfs(root.right, root.val, r);
+    }
+}
+```
+
 ### **...**
 
 ```