feat: add python and java solutions to lcof question

添加《剑指 Offer》题解：面试题36. 二叉搜索树与双向链表

Author: yanglbme

lcof/面试题36. 二叉搜索树与双向链表/README.md

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+# [面试题36. 二叉搜索树与双向链表](https://leetcode-cn.com/problems/er-cha-sou-suo-shu-yu-shuang-xiang-lian-biao-lcof/)
+
+## 题目描述
+<!-- 这里写题目描述 -->
+输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的循环双向链表。要求不能创建任何新的节点，只能调整树中节点指针的指向。
+
+为了让您更好地理解问题，以下面的二叉搜索树为例：
+
+![](https://assets.leetcode.com/uploads/2018/10/12/bstdlloriginalbst.png)
+
+我们希望将这个二叉搜索树转化为双向循环链表。链表中的每个节点都有一个前驱和后继指针。对于双向循环链表，第一个节点的前驱是最后一个节点，最后一个节点的后继是第一个节点。
+
+下图展示了上面的二叉搜索树转化成的链表。“head” 表示指向链表中有最小元素的节点。
+
+![](https://assets.leetcode.com/uploads/2018/10/12/bstdllreturndll.png)
+
+特别地，我们希望可以就地完成转换操作。当转化完成以后，树中节点的左指针需要指向前驱，树中节点的右指针需要指向后继。还需要返回链表中的第一个节点的指针。
+
+**注意**：本题与主站 426 题相同：https://leetcode-cn.com/problems/convert-binary-search-tree-to-sorted-doubly-linked-list/
+
+**注意**：此题对比原题有改动。
+
+## 解法
+<!-- 这里可写通用的实现逻辑 -->
+二叉搜索树中序遍历得到有序序列；根结点指向左子树最后一个结点，根结点也指向右子树第一个结点。
+
+利用虚拟头结点递归遍历求解。
+
+### Python3
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+"""
+
+
+class Solution:
+    def treeToDoublyList(self, root: 'Node') -> 'Node':
+        def dfs(cur):
+            if not cur:
+                return
+            dfs(cur.left)
+            cur.left = self.lastNode
+            if self.lastNode:
+                self.lastNode.right = cur
+            self.lastNode = cur
+            dfs(cur.right)
+
+        if not root:
+            return root
+        self.lastNode = head = Node(-1)
+        dfs(root)
+        head = head.right
+        head.left = self.lastNode
+        self.lastNode.right = head
+        return head
+
+```
+
+### Java
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+/*
+// Definition for a Node.
+class Node {
+    public int val;
+    public Node left;
+    public Node right;
+
+    public Node() {}
+
+    public Node(int _val) {
+        val = _val;
+    }
+
+    public Node(int _val,Node _left,Node _right) {
+        val = _val;
+        left = _left;
+        right = _right;
+    }
+};
+*/
+class Solution {
+    private Node lastNode;
+    public Node treeToDoublyList(Node root) {
+        if (root == null) {
+            return root;
+        }
+        lastNode = new Node(-1);
+        Node head = lastNode;
+        dfs(root);
+        head = head.right;
+        head.left = lastNode;
+        lastNode.right = head;
+        return head;
+    }
+
+    private void dfs(Node cur) {
+        if (cur == null) {
+            return;
+        }
+        dfs(cur.left);
+        cur.left = lastNode;
+        if (lastNode != null) {
+            lastNode.right = cur;
+        }
+        lastNode = cur;
+        dfs(cur.right);
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题36. 二叉搜索树与双向链表/Solution.java

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+/*
+// Definition for a Node.
+class Node {
+    public int val;
+    public Node left;
+    public Node right;
+
+    public Node() {}
+
+    public Node(int _val) {
+        val = _val;
+    }
+
+    public Node(int _val,Node _left,Node _right) {
+        val = _val;
+        left = _left;
+        right = _right;
+    }
+};
+*/
+class Solution {
+    private Node lastNode;
+    public Node treeToDoublyList(Node root) {
+        if (root == null) {
+            return root;
+        }
+        lastNode = new Node(-1);
+        Node head = lastNode;
+        dfs(root);
+        head = head.right;
+        head.left = lastNode;
+        lastNode.right = head;
+        return head;
+    }
+
+    private void dfs(Node cur) {
+        if (cur == null) {
+            return;
+        }
+        dfs(cur.left);
+        cur.left = lastNode;
+        if (lastNode != null) {
+            lastNode.right = cur;
+        }
+        lastNode = cur;
+        dfs(cur.right);
+    }
+}

lcof/面试题36. 二叉搜索树与双向链表/Solution.py

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+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+"""
+
+
+class Solution:
+    def treeToDoublyList(self, root: 'Node') -> 'Node':
+        def dfs(cur):
+            if not cur:
+                return
+            dfs(cur.left)
+            cur.left = self.lastNode
+            if self.lastNode:
+                self.lastNode.right = cur
+            self.lastNode = cur
+            dfs(cur.right)
+
+        if not root:
+            return root
+        self.lastNode = head = Node(-1)
+        dfs(root)
+        head = head.right
+        head.left = self.lastNode
+        self.lastNode.right = head
+        return head