From 41224f61bd37979847ee6258cda1282aa227befe Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Thu, 1 Jun 2023 22:50:49 -0500
Subject: [PATCH] =?UTF-8?q?Update=200096.=E4=B8=8D=E5=90=8C=E7=9A=84?=
 =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0096.不同的二叉搜索树.md | 13 +++++++------
 1 file changed, 7 insertions(+), 6 deletions(-)

diff --git a/problems/0096.不同的二叉搜索树.md b/problems/0096.不同的二叉搜索树.md
index 9f41906d..368a5747 100644
--- a/problems/0096.不同的二叉搜索树.md
+++ b/problems/0096.不同的二叉搜索树.md
@@ -197,12 +197,13 @@ class Solution {
 ```python
 class Solution:
     def numTrees(self, n: int) -> int:
-        dp = [0] * (n + 1)
-        dp[0], dp[1] = 1, 1
-        for i in range(2, n + 1):
-            for j in range(1, i + 1):
-                dp[i] += dp[j - 1] * dp[i - j]
-        return dp[-1]
+        dp = [0] * (n + 1)  # 创建一个长度为n+1的数组，初始化为0
+        dp[0] = 1  # 当n为0时，只有一种情况，即空树，所以dp[0] = 1
+        for i in range(1, n + 1):  # 遍历从1到n的每个数字
+            for j in range(1, i + 1):  # 对于每个数字i，计算以i为根节点的二叉搜索树的数量
+                dp[i] += dp[j - 1] * dp[i - j]  # 利用动态规划的思想，累加左子树和右子树的组合数量
+        return dp[n]  # 返回以1到n为节点的二叉搜索树的总数量
+
 ```
 
 ### Go