From 37d12a168cfa59e0ae44203f5aa8234d588f23fd Mon Sep 17 00:00:00 2001
From: ZerenZhang2022 <118794589+ZerenZhang2022@users.noreply.github.com>
Date: Wed, 22 Mar 2023 18:22:15 -0400
Subject: [PATCH 1/2] =?UTF-8?q?Update=200530.=E4=BA=8C=E5=8F=89=E6=90=9C?=
 =?UTF-8?q?=E7=B4=A2=E6=A0=91=E7=9A=84=E6=9C=80=E5=B0=8F=E7=BB=9D=E5=AF=B9?=
 =?UTF-8?q?=E5=B7=AE.md?=
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Content-Type: text/plain; charset=UTF-8
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增加python - 双指针法，不用数组 (同Carl写法) - 更快
---
 .../0530.二叉搜索树的最小绝对差.md | 19 +++++++++++++++++++
 1 file changed, 19 insertions(+)

diff --git a/problems/0530.二叉搜索树的最小绝对差.md b/problems/0530.二叉搜索树的最小绝对差.md
index 203add39..fa1430de 100644
--- a/problems/0530.二叉搜索树的最小绝对差.md
+++ b/problems/0530.二叉搜索树的最小绝对差.md
@@ -221,8 +221,27 @@ class Solution:
         for i in range(len(res)-1):  // 统计有序数组的最小差值
             r = min(abs(res[i]-res[i+1]),r)
         return r
+        
+        
+class Solution: # 双指针法，不用数组 (同Carl写法) - 更快
+    def getMinimumDifference(self, root: Optional[TreeNode]) -> int:
+        global pre,minval
+        pre = None
+        minval = 10**5
+        self.traversal(root)
+        return minval
+
+    def traversal(self,root):
+        global pre,minval
+        if not root: return None
+        self.traversal(root.left)
+        if pre and root.val-pre.val<minval: 
+            minval = root.val-pre.val
+        pre = root
+        self.traversal(root.right)            
 ```
 
+
 迭代法-中序遍历
 ```python 
 class Solution:

From c73c1a2b107780ce0aeb78c578112d065b887aa4 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BA=A6=E8=B6=85=E7=A6=B9?= <1677425105@qq.com>
Date: Sun, 26 Mar 2023 18:36:13 +0800
Subject: [PATCH 2/2] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E7=90=86?=
 =?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-1.md?=
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Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/背包理论基础01背包-1.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md
index ff0b6aba..c45fc3d3 100644
--- a/problems/背包理论基础01背包-1.md
+++ b/problems/背包理论基础01背包-1.md
@@ -87,7 +87,7 @@ leetcode上没有纯01背包的问题，都是01背包应用方面的题目，
 
 那么可以有两个方向推出来dp[i][j]，
 
-* **不放物品i**：由dp[i - 1][j]推出，即背包容量为j，里面不放物品i的最大价值，此时dp[i][j]就是dp[i - 1][j]。(其实就是当物品i的重量大于背包j的重量时，物品i无法放进背包中，所以被背包内的价值依然和前面相同。)
+* **不放物品i**：由dp[i - 1][j]推出，即背包容量为j，里面不放物品i的最大价值，此时dp[i][j]就是dp[i - 1][j]。(其实就是当物品i的重量大于背包j的重量时，物品i无法放进背包中，所以背包内的价值依然和前面相同。)
 * **放物品i**：由dp[i - 1][j - weight[i]]推出，dp[i - 1][j - weight[i]] 为背包容量为j - weight[i]的时候不放物品i的最大价值，那么dp[i - 1][j - weight[i]] + value[i] （物品i的价值），就是背包放物品i得到的最大价值
 
 所以递归公式： dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);