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feat: update leetcode solutions: No.0235, No.0236
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‎README.md‎

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### 二叉树
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1. [二叉树的最近公共祖先](./lcof/面试题68%20-%20II.%20二叉树的最近公共祖先/README.md)
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1. [二叉搜索树的最近公共祖先](./lcof/面试题68%20-%20I.%20二叉搜索树的最近公共祖先/README.md)
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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)
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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)
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### 数学
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‎README_EN.md‎

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@@ -47,6 +47,10 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
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### Binary Tree
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1. [Lowest Common Ancestor of a Binary Tree](./solution/0200-0299/0236.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree/README_EN.md)
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1. [Lowest Common Ancestor of a Binary Search Tree](./solution/0200-0299/0235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree/README_EN.md)
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### Math
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1. [Set Mismatch](./solution/0600-0699/0645.Set%20Mismatch/README_EN.md)

‎solution/0200-0299/0235.Lowest Common Ancestor of a Binary Search Tree/README.md‎

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<p>例如，给定如下二叉搜索树:&nbsp; root =&nbsp;[6,2,8,0,4,7,9,null,null,3,5]</p>
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<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/14/binarysearchtree_improved.png" style="height: 190px; width: 200px;"></p>
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<p>&nbsp;</p>
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![](./images/binarysearchtree_improved.png)
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<p><strong>示例 1:</strong></p>
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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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从上到下搜索，找到第一个值位于 `[p, q]` 之间的结点即可。既可以用迭代实现，也可以用递归实现。
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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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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
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if p == q:
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return p
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while root:
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if root.val < p.val and root.val < q.val:
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root = root.right
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elif root.val > p.val and root.val > q.val:
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root = root.left
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else:
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return root
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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(int x) { val = x; }
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* }
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*/
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class Solution {
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (p == q) {
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return p;
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}
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while (root != null) {
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if (root.val < p.val && root.val < q.val) {
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root = root.right;
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} else if (root.val > p.val && root.val > q.val) {
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root = root.left;
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} else {
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return root;
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}
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}
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return null;
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}
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}
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```
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### **...**

‎solution/0200-0299/0235.Lowest Common Ancestor of a Binary Search Tree/README_EN.md‎

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<p>Given binary search tree:&nbsp; root =&nbsp;[6,2,8,0,4,7,9,null,null,3,5]</p>
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<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png" style="width: 200px; height: 190px;" />
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<p>&nbsp;</p>
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![](./images/binarysearchtree_improved.png)
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<p><strong>Example 1:</strong></p>
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<ul>
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<li>All of the nodes&#39; values will be unique.</li>
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<li>p and q are different and both values will&nbsp;exist in the BST.</li>
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</ul>
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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, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
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if p == q:
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return p
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while root:
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if root.val < p.val and root.val < q.val:
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root = root.right
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elif root.val > p.val and root.val > q.val:
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root = root.left
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else:
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return root
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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(int x) { val = x; }
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* }
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*/
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class Solution {
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (p == q) {
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return p;
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}
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while (root != null) {
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if (root.val < p.val && root.val < q.val) {
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root = root.right;
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} else if (root.val > p.val && root.val > q.val) {
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root = root.left;
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} else {
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return root;
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}
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}
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return null;
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}
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}
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```
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### **...**

‎solution/0200-0299/0235.Lowest Common Ancestor of a Binary Search Tree/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(int x) { val = x; }
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* }
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*/
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class Solution {
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (root == null || root == p || root == q) {
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return root;
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if (p == q) {
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return p;
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}
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TreeNode leftNode = lowestCommonAncestor(root.left, p, q);
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TreeNode rightNode = lowestCommonAncestor(root.right, p, q);
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if (leftNode != null && rightNode != null) {
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return root;
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while (root != null) {
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if (root.val < p.val && root.val < q.val) {
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root = root.right;
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} else if (root.val > p.val && root.val > q.val) {
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root = root.left;
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} else {
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return root;
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}
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}
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return leftNode != null ? leftNode : (rightNode != null) ? rightNode : null;
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return null;
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}
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}

‎solution/0200-0299/0235.Lowest Common Ancestor of a Binary Search Tree/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, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
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if p == q:
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return p
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while root:
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if root.val < p.val and root.val < q.val:
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root = root.right
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elif root.val > p.val and root.val > q.val:
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root = root.left
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else:
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return root

‎solution/0200-0299/0235.Lowest Common Ancestor of a Binary Search Tree/images/binarysearchtree_improved.png‎

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‎solution/0200-0299/0236.Lowest Common Ancestor of a Binary Tree/README.md‎

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<p>例如，给定如下二叉树:&nbsp; root =&nbsp;[3,5,1,6,2,0,8,null,null,7,4]</p>
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<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/15/binarytree.png" style="height: 190px; width: 200px;"></p>
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<p>&nbsp;</p>
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![](./images/binarytree.png)
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<p><strong>示例 1:</strong></p>
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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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根据“**最近公共祖先**”的定义，若 root 是 p, q 的最近公共祖先 ，则只可能为以下情况之一：
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- 如果 p 和 q 分别是 root 的左右节点，那么 root 就是我们要找的最近公共祖先；
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- 如果 p 和 q 都是 root 的左节点，那么返回 `lowestCommonAncestor(root.left, p, q)`；
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- 如果 p 和 q 都是 root 的右节点，那么返回 `lowestCommonAncestor(root.right, p, q)`。
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**边界条件讨论**：
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- 如果 root 为 null，则说明我们已经找到最底了，返回 null 表示没找到；
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- 如果 root 与 p 相等或者与 q 相等，则返回 root；
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- 如果左子树没找到，递归函数返回 null，证明 p 和 q 同在 root 的右侧，那么最终的公共祖先就是右子树找到的结点；
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- 如果右子树没找到，递归函数返回 null，证明 p 和 q 同在 root 的左侧，那么最终的公共祖先就是左子树找到的结点。
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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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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
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if root is None or root == p or root == q:
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return root
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left = self.lowestCommonAncestor(root.left, p, q)
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right = self.lowestCommonAncestor(root.right, p, q)
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return right if left is None else (left if right is None else root)
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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(int x) { val = x; }
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* }
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*/
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class Solution {
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (root == null || root == p || root == q) {
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return root;
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}
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TreeNode left = lowestCommonAncestor(root.left, p, q);
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TreeNode right = lowestCommonAncestor(root.right, p, q);
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return left == null ? right : (right == null ? left : root);
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}
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}
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```
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### **...**

‎solution/0200-0299/0236.Lowest Common Ancestor of a Binary Tree/README_EN.md‎

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<p>Given the following binary tree:&nbsp; root =&nbsp;[3,5,1,6,2,0,8,null,null,7,4]</p>
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<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;" />
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<p>&nbsp;</p>
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![](./images/binarytree.png)
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<p><strong>Example 1:</strong></p>
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<ul>
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<li>All of the nodes&#39; values will be unique.</li>
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<li>p and q are different and both values will&nbsp;exist in the binary tree.</li>
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</ul>
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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, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
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if root is None or root == p or root == q:
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return root
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left = self.lowestCommonAncestor(root.left, p, q)
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right = self.lowestCommonAncestor(root.right, p, q)
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return right if left is None else (left if right is None else root)
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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(int x) { val = x; }
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* }
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*/
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class Solution {
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (root == null || root == p || root == q) {
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return root;
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}
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TreeNode left = lowestCommonAncestor(root.left, p, q);
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TreeNode right = lowestCommonAncestor(root.right, p, q);
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return left == null ? right : (right == null ? left : root);
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}
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}
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```
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### **...**

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