feat: add python and java solutions to lcci problem: No.17.12

See https://leetcode-cn.com/problems/binode-lcci/

Author: yanglbme

README.md

@@ -100,6 +100,7 @@
 1. [二叉搜索树的后序遍历序列](/lcof/面试题33.%20二叉搜索树的后序遍历序列/README.md)
 1. [二叉树的最近公共祖先](/solution/0200-0299/0235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree/README.md)
 1. [二叉搜索树的最近公共祖先](/solution/0200-0299/0236.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree/README.md)
+1. [将二叉搜索树转换为单链表](/lcci/17.12.BiNode/README.md)
 
 ### 数学
 

README_EN.md

@@ -96,6 +96,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 1. [Construct Binary Tree from Inorder and Postorder Traversal](/solution/0100-0199/0106.Construct%20Binary%20Tree%20from%20Inorder%20and%20Postorder%20Traversal/README_EN.md)
 1. [Lowest Common Ancestor of a Binary Tree](/solution/0200-0299/0236.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree/README_EN.md)
 1. [Lowest Common Ancestor of a Binary Search Tree](/solution/0200-0299/0235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree/README_EN.md)
+1. [BiNode](/lcci/17.12.BiNode/README_EN.md)
 
 ### Math
 

lcci/17.12.BiNode/README.md

@@ -29,22 +29,73 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+递归将左子树、右子树转换为左、右链表 left 和 right。然后将左链表 left 的最后一个结点的 right 指针指向 root，root 的 right 指针指向右链表 right，并将 root 的 left 指针值为空。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def convertBiNode(self, root: TreeNode) -> TreeNode:
+        if root is None:
+            return None
+        left = self.convertBiNode(root.left)
+        right = self.convertBiNode(root.right)
+        if left is None:
+            root.right = right
+            return root
+        res = left
+        while left and left.right:
+            left = left.right
+        left.right = root
+        root.right = right
+        root.left = None
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public TreeNode convertBiNode(TreeNode root) {
+        if (root == null) return null;
+        TreeNode left = convertBiNode(root.left);
+        TreeNode right = convertBiNode(root.right);
+        if (left == null) {
+            root.right = right;
+            return root;
+        }
+        TreeNode res = left;
+        while (left != null && left.right != null) {
+            left = left.right;
+        }
+        left.right = root;
+        root.right = right;
+        root.left = null;
+        return res;
+    }
+}
 ```
 
 ### **...**

lcci/17.12.BiNode/README_EN.md

@@ -35,13 +35,62 @@
 ### **Python3**
 
 ```python
-
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def convertBiNode(self, root: TreeNode) -> TreeNode:
+        if root is None:
+            return None
+        left = self.convertBiNode(root.left)
+        right = self.convertBiNode(root.right)
+        if left is None:
+            root.right = right
+            return root
+        res = left
+        while left and left.right:
+            left = left.right
+        left.right = root
+        root.right = right
+        root.left = None
+        return res
 ```
 
 ### **Java**
 
 ```java
-
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public TreeNode convertBiNode(TreeNode root) {
+        if (root == null) return null;
+        TreeNode left = convertBiNode(root.left);
+        TreeNode right = convertBiNode(root.right);
+        if (left == null) {
+            root.right = right;
+            return root;
+        }
+        TreeNode res = left;
+        while (left != null && left.right != null) {
+            left = left.right;
+        }
+        left.right = root;
+        root.right = right;
+        root.left = null;
+        return res;
+    }
+}
 ```
 
 ### **...**

lcci/17.12.BiNode/Solution.java

@@ -0,0 +1,28 @@
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public TreeNode convertBiNode(TreeNode root) {
+        if (root == null) return null;
+        TreeNode left = convertBiNode(root.left);
+        TreeNode right = convertBiNode(root.right);
+        if (left == null) {
+            root.right = right;
+            return root;
+        }
+        TreeNode res = left;
+        while (left != null && left.right != null) {
+            left = left.right;
+        }
+        left.right = root;
+        root.right = right;
+        root.left = null;
+        return res;
+    }
+}

lcci/17.12.BiNode/Solution.py

@@ -0,0 +1,23 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def convertBiNode(self, root: TreeNode) -> TreeNode:
+        if root is None:
+            return None
+        left = self.convertBiNode(root.left)
+        right = self.convertBiNode(root.right)
+        if left is None:
+            root.right = right
+            return root
+        res = left
+        while left and left.right:
+            left = left.right
+        left.right = root
+        root.right = right
+        root.left = None
+        return res