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feat: add solutions to lc problem: No.0230. Kth Smallest Element in a BST
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‎solution/0200-0299/0230.Kth Smallest Element in a BST/README.md

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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def kthSmallest(self, root: TreeNode, k: int) -> int:
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def inorder(root):
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if root is None:
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return
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inorder(root.left)
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self.k -= 1
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if self.k == 0:
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self.res = root.val
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return
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inorder(root.right)
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self.k = k
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inorder(root)
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return self.res
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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private int k;
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private int res;
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public int kthSmallest(TreeNode root, int k) {
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this.k = k;
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inorder(root);
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return res;
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}
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private void inorder(TreeNode root) {
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if (root == null) {
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return;
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}
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inorder(root.left);
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if (--k == 0) {
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res = root.val;
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return;
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}
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inorder(root.right);
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}
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}
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```
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### **...**

‎solution/0200-0299/0230.Kth Smallest Element in a BST/README_EN.md

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### **Python3**
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def kthSmallest(self, root: TreeNode, k: int) -> int:
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def inorder(root):
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if root is None:
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return
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inorder(root.left)
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self.k -= 1
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if self.k == 0:
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self.res = root.val
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return
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inorder(root.right)
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self.k = k
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inorder(root)
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return self.res
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```
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### **Java**
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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private int k;
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private int res;
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public int kthSmallest(TreeNode root, int k) {
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this.k = k;
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inorder(root);
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return res;
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}
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private void inorder(TreeNode root) {
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if (root == null) {
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return;
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}
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inorder(root.left);
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if (--k == 0) {
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res = root.val;
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return;
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}
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inorder(root.right);
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}
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}
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```
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### **...**

‎solution/0200-0299/0230.Kth Smallest Element in a BST/Solution.java

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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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private int k;
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private int res;
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public int kthSmallest(TreeNode root, int k) {
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Deque<TreeNode> stack = new ArrayDeque<>();
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while (!stack.isEmpty() || root != null) {
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while (root != null) {
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stack.push(root);
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root = root.left;
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}
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root = stack.pop();
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if (--k == 0) {
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return root.val;
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}
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root = root.right;
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this.k = k;
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inorder(root);
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return res;
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}
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private void inorder(TreeNode root) {
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if (root == null) {
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return;
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}
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inorder(root.left);
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if (--k == 0) {
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res = root.val;
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return;
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}
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return 0;
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inorder(root.right);
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}
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}
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}

‎solution/0200-0299/0230.Kth Smallest Element in a BST/Solution.py

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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def kthSmallest(self, root: TreeNode, k: int) -> int:
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def inorder(root):
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if root is None:
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return
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inorder(root.left)
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self.k -= 1
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if self.k == 0:
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self.res = root.val
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return
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inorder(root.right)
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self.k = k
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inorder(root)
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return self.res

‎solution/1900-1999/1901.Find a Peak Element II/README.md

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# [1901. Find a Peak Element II](https://leetcode-cn.com/problems/find-a-peak-element-ii)
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[English Version](/solution/1900-1999/1901.Find%20a%20Peak%20Element%20II/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>A <strong>peak</strong> element in a 2D grid is an element that is <strong>strictly greater</strong> than all of its <strong>adjacent </strong>neighbors to the left, right, top, and bottom.</p>
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<p>Given a <strong>0-indexed</strong> <code>m x n</code> matrix <code>mat</code> where <strong>no two adjacent cells are equal</strong>, find <strong>any</strong> peak element <code>mat[i][j]</code> and return <em>the length 2 array </em><code>[i,j]</code>.</p>
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<p>You may assume that the entire matrix is surrounded by an <strong>outer perimeter</strong> with the value <code>-1</code> in each cell.</p>
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<p>You must write an algorithm that runs in <code>O(m log(n))</code> or <code>O(n log(m))</code> time.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1901.Find%20a%20Peak%20Element%20II/images/1.png" style="width: 206px; height: 209px;" /></p>
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<pre>
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<strong>Input:</strong> mat = [[1,4],[3,2]]
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<strong>Output:</strong> [0,1]
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<strong>Explanation:</strong>&nbsp;Both 3 and 4 are peak elements so [1,0] and [0,1] are both acceptable answers.
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</pre>
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<p><strong>Example 2:</strong></p>
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<p><strong><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1901.Find%20a%20Peak%20Element%20II/images/3.png" style="width: 254px; height: 257px;" /></strong></p>
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<pre>
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<strong>Input:</strong> mat = [[10,20,15],[21,30,14],[7,16,32]]
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<strong>Output:</strong> [1,1]
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<strong>Explanation:</strong>&nbsp;Both 30 and 32 are peak elements so [1,1] and [2,2] are both acceptable answers.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>m == mat.length</code></li>
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<li><code>n == mat[i].length</code></li>
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<li><code>1 &lt;= m, n &lt;= 500</code></li>
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<li><code>1 &lt;= mat[i][j] &lt;= 10<sup>5</sup></code></li>
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<li>No two adjacent cells are equal.</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1900-1999/1901.Find a Peak Element II/README_EN.md

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# [1901. Find a Peak Element II](https://leetcode.com/problems/find-a-peak-element-ii)
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[中文文档](/solution/1900-1999/1901.Find%20a%20Peak%20Element%20II/README.md)
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## Description
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<p>A <strong>peak</strong> element in a 2D grid is an element that is <strong>strictly greater</strong> than all of its <strong>adjacent </strong>neighbors to the left, right, top, and bottom.</p>
8+
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<p>Given a <strong>0-indexed</strong> <code>m x n</code> matrix <code>mat</code> where <strong>no two adjacent cells are equal</strong>, find <strong>any</strong> peak element <code>mat[i][j]</code> and return <em>the length 2 array </em><code>[i,j]</code>.</p>
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<p>You may assume that the entire matrix is surrounded by an <strong>outer perimeter</strong> with the value <code>-1</code> in each cell.</p>
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<p>You must write an algorithm that runs in <code>O(m log(n))</code> or <code>O(n log(m))</code> time.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1901.Find%20a%20Peak%20Element%20II/images/1.png" style="width: 206px; height: 209px;" /></p>
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<pre>
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<strong>Input:</strong> mat = [[1,4],[3,2]]
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<strong>Output:</strong> [0,1]
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<strong>Explanation:</strong>&nbsp;Both 3 and 4 are peak elements so [1,0] and [0,1] are both acceptable answers.
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</pre>
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<p><strong>Example 2:</strong></p>
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<p><strong><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1901.Find%20a%20Peak%20Element%20II/images/3.png" style="width: 254px; height: 257px;" /></strong></p>
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<pre>
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<strong>Input:</strong> mat = [[10,20,15],[21,30,14],[7,16,32]]
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<strong>Output:</strong> [1,1]
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<strong>Explanation:</strong>&nbsp;Both 30 and 32 are peak elements so [1,1] and [2,2] are both acceptable answers.
34+
</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>m == mat.length</code></li>
41+
<li><code>n == mat[i].length</code></li>
42+
<li><code>1 &lt;= m, n &lt;= 500</code></li>
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<li><code>1 &lt;= mat[i][j] &lt;= 10<sup>5</sup></code></li>
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<li>No two adjacent cells are equal.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1900-1999/1901.Find a Peak Element II/images/1.png

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‎solution/1900-1999/1901.Find a Peak Element II/images/3.png

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