feat: add solutions to lcof2 problem: No.052 (doocs#1466)

No.052.展开二叉搜索树

Author: yanglbme

lcof2/剑指 Offer II 052. 展平二叉搜索树/README.md

@@ -78,6 +78,33 @@ class Solution:
         return head
 ```
 
+```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 increasingBST(self, root: TreeNode) -> TreeNode:
+        def dfs(root: TreeNode):
+            if root is None:
+                return
+
+            dfs(root.left)
+
+            nonlocal cur
+            cur.right = root
+            root.left = None
+            cur = cur.right
+
+            dfs(root.right)
+
+        cur = dummy = TreeNode()
+        dfs(root)
+        return dummy.right
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -123,6 +150,45 @@ 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 {
+    private TreeNode cur;
+
+    public TreeNode increasingBST(TreeNode root) {
+        TreeNode dummy = new TreeNode();
+        cur = dummy;
+        dfs(root);
+        return dummy.right;
+    }
+
+    private void dfs(TreeNode root) {
+        if (root == null) {
+            return;
+        }
+        dfs(root.left);
+        cur.right = root;
+        root.left = null;
+        cur = cur.right;
+        dfs(root.right);
+    }
+}
+```
+
 ### **Go**
 
 ```go
@@ -158,6 +224,34 @@ func increasingBST(root *TreeNode) *TreeNode {
 }
 ```
 
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func increasingBST(root *TreeNode) *TreeNode {
+	dummy := &TreeNode{}
+	cur := dummy
+	var dfs func(*TreeNode)
+	dfs = func(root *TreeNode) {
+		if root == nil {
+			return
+		}
+		dfs(root.Left)
+		root.Left = nil
+		cur.Right = root
+		cur = root
+		dfs(root.Right)
+	}
+	dfs(root)
+	return dummy.Right
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -199,6 +293,39 @@ 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:
+    TreeNode* increasingBST(TreeNode* root) {
+        TreeNode* dummy = new TreeNode();
+        TreeNode* cur = dummy;
+        function<void(TreeNode*)> dfs = [&](TreeNode* root) {
+            if (!root) {
+                return;
+            }
+            dfs(root->left);
+            cur->right = root;
+            root->left = nullptr;
+            cur = cur->right;
+            dfs(root->right);
+        };
+        dfs(root);
+        return dummy->right;
+    }
+};
+```
+
 ### **TypeScript**
 
 ```ts