From 2b33b45170c22eb4b900bc85ff04ad5ce0a5c2e2 Mon Sep 17 00:00:00 2001
From: Guanzhong Pan <gpan20020205@gmail.com>
Date: Sat, 22 Jan 2022 10:39:07 +0000
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200096.=E4=B8=8D=E5=90=8C?=
 =?UTF-8?q?=E7=9A=84=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md=20C?=
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---
 problems/0096.不同的二叉搜索树.md | 27 +++++++++++++++++++++++
 1 file changed, 27 insertions(+)

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 d4b8d024..d6f7aca7 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
@@ -227,7 +227,34 @@ const numTrees =(n) => {
 };
 ```
 
+C:
+```c
+//开辟dp数组
+int *initDP(int n) {
+    int *dp = (int *)malloc(sizeof(int) * (n + 1));
+    int i;
+    for(i = 0; i <= n; ++i) 
+        dp[i] = 0;
+    return dp;
+}
 
+int numTrees(int n){
+    //开辟dp数组
+    int *dp = initDP(n);
+    //将dp[0]设为1
+    dp[0] = 1;
+
+    int i, j;
+    for(i = 1; i <= n; ++i) {
+        for(j = 1; j <= i; ++j) {
+            //递推公式：dp[i] = d[i] + 根为j时左子树种类个数 * 根为j时右子树种类个数
+            dp[i] += dp[j - 1] * dp[i - j];
+        }
+    }
+
+    return dp[n];
+}
+```
 
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>