feat: #236 #238

README.md

@@ -120,6 +120,8 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - [201.bitwise-and-of-numbers-range](./problems/201.bitwise-and-of-numbers-range.md)
 - 🆕 [208.implement-trie-prefix-tree](./problems/208.implement-trie-prefix-tree.md)
 - 🖊 [209.minimum-size-subarray-sum](./problems/209.minimum-size-subarray-sum.md)
+- 🆕 [236.lowest-common-ancestor-of-a-binary-tree](./problems/236.lowest-common-ancestor-of-a-binary-tree.md)
+- 🆕 [238.product-of-array-except-self](./problems/238.product-of-array-except-self.md)
 - [240.search-a-2-d-matrix-ii](./problems/240.search-a-2-d-matrix-ii.md)
 - 🖊 [279.perfect-squares](./problems/279.perfect-squares.md)
 - [322.coin-change](./problems/322.coin-change.md)

assets/drawio/238.product-of-array-except-self.drawio

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\ No newline at end of file

236.lowest-common-ancestor-of-a-binary-tree.js → problems/236.lowest-common-ancestor-of-a-binary-tree.md

+## 题目地址
+
+https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
+
+## 题目描述
+
+```
+Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
+
+According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
+
+Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]
+
+```
+![236.lowest-common-ancestor-of-a-binary-tree](../assets/problems/236.lowest-common-ancestor-of-a-binary-tree-1.png)
+
+```
+Example 1:
+
+Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
+Output: 3
+Explanation: The LCA of nodes 5 and 1 is 3.
+Example 2:
+
+Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
+Output: 5
+Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
+ 
+
+Note:
+
+All of the nodes' values will be unique.
+p and q are different and both values will exist in the binary tree.
+```
+
+## 思路
+
+这道题目是求解二叉树中，两个给定节点的最近的公共祖先。是一道非常经典的二叉树题目。
+
+我们之前说过树是一种递归的数据结构，因此使用递归方法解决二叉树问题从写法上来看是最简单的，这道题目也不例外。
+
+用递归的思路去思考树是一种非常重要的能力。
+
+
+如果大家这样去思考的话，问题就会得到简化，我们的目标就是分别在左右子树进行查找p和q。 如果p没有在左子树，那么它一定在右子树（题目限定p一定在树中），
+反之亦然。
+
+对于具体的代码而言就是，我们假设这个树就一个结构，然后尝试去解决，然后在适当地方去递归自身即可。 如下图所示：
+
+![236.lowest-common-ancestor-of-a-binary-tree-2](../assets/problems/236.lowest-common-ancestor-of-a-binary-tree-2.png)
+
+我们来看下核心代码：
+
+```js
+  // 如果我们找到了p，直接进行返回，那如果下面就是q呢？ 其实这没有影响，但是还是要多考虑一下
+  if (!root || root === p || root === q) return root;
+  const left = lowestCommonAncestor(root.left, p, q); // 去左边找，我们期望返回找到的节点
+  const right = lowestCommonAncestor(root.right, p, q);// 去右边找，我们期望返回找到的节点
+  if (!left) return right; // 左子树找不到，返回右子树
+  if (!right) return left; // 右子树找不到，返回左子树
+  return root; // 左右子树分别有一个，则返回root
+
+```
+
+> 如果没有明白的话，请多花时间消化一下
+
+## 关键点解析
+
+- 用递归的思路去思考树
+
+## 代码
+
+```js
+
+
 /*
  * @lc app=leetcode id=236 lang=javascript
  *
@@ -70,3 +145,5 @@ var lowestCommonAncestor = function(root, p, q) {
   if (!right) return left; // 右子树找不到，返回左子树
   return root; // 左右子树分别有一个，则返回root
 };
+```
+

238.product-of-array-except-self.js → problems/238.product-of-array-except-self.md

+## 题目地址
+
+https://leetcode.com/problems/product-of-array-except-self/description/
+
+## 题目描述
+
+```
+Given an array nums of n integers where n > 1,  return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
+
+Example:
+
+Input:  [1,2,3,4]
+Output: [24,12,8,6]
+Note: Please solve it without division and in O(n).
+
+Follow up:
+Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)
+
+
+```
+
+## 思路
+
+这道题的意思是给定一个数组，返回一个新的数组，这个数组每一项都是其他项的乘积。 
+符合直觉的思路是两层循环，时间复杂度是O(n),但是题目要求`Please solve it without division and in O(n)`。
+
+因此我们需要换一种思路，由于输出的每一项都需要用到别的元素，因此一次遍历是绝对不行的。 
+考虑我们先进行一次遍历， 然后维护一个数组，第i项代表前i个元素（不包括i）的乘积。
+然后我们反向遍历一次，然后维护另一个数组，同样是第i项代表前i个元素（不包括i）的乘积。
+
+![238.product-of-array-except-self](../assets/problems/238.product-of-array-except-self.png)
+
+有意思的是第一个数组和第二个数组的反转（reverse）做乘法（有点像向量运算）就是我们想要的运算。
+
+其实我们进一步观察，我们不需要真的创建第二个数组（第二个数组只是做中间运算使用），而是直接修改第一个数组即可。
+
+## 关键点解析
+
+- 两次遍历， 一次正向，一次反向。
+- 维护一个数组，第i项代表前i个元素（不包括i）的乘积
+
+## 代码
+
+```js
+
 /*
  * @lc app=leetcode id=238 lang=javascript
  *
@@ -54,3 +99,4 @@ var productExceptSelf = function(nums) {
   }
   return ret;
 };
+```