From a4c817373e80c65ca773a48155e833b0ec21f719 Mon Sep 17 00:00:00 2001
From: Stone Lyu <mengyulyu@gmail.com>
Date: Tue, 24 Dec 2019 15:26:47 +0800
Subject: [PATCH] feat: #230 add java solution(#249)
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* correct typos

* Add translations

* 新增HTTP/1.0, HTTP/1.1， HTTP/2的对比

* 增加HTTP/1.1和HTTP/2的性能对比的Demo

* 翻译大部分基础数据结构，但是REACT部分没有翻译

* update description of `HTTP/1.0`

* add WIP tag, update some translation details

* change 'fundamental' to 'basic'

* remove some trivial details about HTTP/1.0

* error correction

* translation for DP

* translation for string problems

* error correction

* correct wrongly typed

* translation for binary-tree-traversal.md

* correct wrongly typed

* update README.en.md

* update description  of Huffman encoding

* update file path of Binary-tree-traversal-en.md.

* correct wrongly typed

* update paths of thinkings

* translation for basic-algorithm.md (Draft)

* translation for bloom-filter.md

* add java solution with inorder traversing

* Update 230.kth-smallest-element-in-a-bst.md

Co-authored-by: lucifer <azl397985856@gmail.com>
---
 problems/230.kth-smallest-element-in-a-bst.md | 37 +++++++++++++++++++
 1 file changed, 37 insertions(+)

diff --git a/problems/230.kth-smallest-element-in-a-bst.md b/problems/230.kth-smallest-element-in-a-bst.md
index 566405c90..b11e07d78 100644
--- a/problems/230.kth-smallest-element-in-a-bst.md
+++ b/problems/230.kth-smallest-element-in-a-bst.md
@@ -56,6 +56,8 @@ What if the BST is modified (insert/delete operations) often and you need to fin
 
 解法一：
 
+JavaScript Code:
+
 ```js
 
 
@@ -107,9 +109,44 @@ var kthSmallest = function(root, k) {
 };
 ```
 
+Java Code:
+
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+private int count = 1;
+private int res;
+
+public int KthSmallest (TreeNode root, int k) {
+    inorder(root, k);
+    return res;
+}
+
+public void inorder (TreeNode root, int k) {
+    if (root == null) return;
+
+    inorder(root.left, k);
+
+    if (count++ == k) {
+        res = root.val;
+        return;
+    }
+    
+    inorder(root.right, k);
+}
+```
 
 解法二：
 
+JavaScript Code:
+
 ```js
 
 /**