From 3947eeaab9d868fbe4665629a1792c1c9cd3a593 Mon Sep 17 00:00:00 2001
From: L1Y1 <108316255+L1Y1@users.noreply.github.com>
Date: Thu, 13 Oct 2022 17:00:01 +0800
Subject: [PATCH 1/2] =?UTF-8?q?Update=200235.=E4=BA=8C=E5=8F=89=E6=90=9C?=
 =?UTF-8?q?=E7=B4=A2=E6=A0=91=E7=9A=84=E6=9C=80=E8=BF=91=E5=85=AC=E5=85=B1?=
 =?UTF-8?q?=E7=A5=96=E5=85=88.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

可以写的更通俗易懂一些
---
 problems/0235.二叉搜索树的最近公共祖先.md | 6 ++----
 1 file changed, 2 insertions(+), 4 deletions(-)

diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md
index 07e55ba5..a8c94ec9 100644
--- a/problems/0235.二叉搜索树的最近公共祖先.md
+++ b/problems/0235.二叉搜索树的最近公共祖先.md
@@ -328,13 +328,11 @@ var lowestCommonAncestor = function(root, p, q) {
     }
     if(root.val>p.val&&root.val>q.val) {
         // 向左子树查询
-        let left = lowestCommonAncestor(root.left,p,q);
-        return left !== null&&left;
+         return root.left = lowestCommonAncestor(root.left,p,q);
     }
     if(root.val<p.val&&root.val<q.val) {
         // 向右子树查询
-        let right = lowestCommonAncestor(root.right,p,q);
-        return right !== null&&right;
+        return root.right = lowestCommonAncestor(root.right,p,q);
     }
     return root;
 };

From 819387081395ada6a866e5ce9ca236319cc2ca00 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=A4=A7=E7=9C=BC=E7=9D=9B=E5=9B=BE=E5=9B=BE?=
 <108316255+L1Y1@users.noreply.github.com>
Date: Sun, 16 Oct 2022 20:34:22 +0800
Subject: [PATCH 2/2] =?UTF-8?q?Update=200216.=E7=BB=84=E5=90=88=E6=80=BB?=
 =?UTF-8?q?=E5=92=8CIII.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

感觉这种解法更符合卡哥视频的讲解。
---
 problems/0216.组合总和III.md | 35 +++++++++++++++++++-------------
 1 file changed, 21 insertions(+), 14 deletions(-)

diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md
index 964facee..4273240c 100644
--- a/problems/0216.组合总和III.md
+++ b/problems/0216.组合总和III.md
@@ -397,24 +397,31 @@ func backTree(n,k,startIndex int,track *[]int,result *[][]int){
  * @return {number[][]}
  */
 var combinationSum3 = function(k, n) {
-    const backtrack = (start) => {
-        const l = path.length;
-        if (l === k) {
-            const sum = path.reduce((a, b) => a + b);
-            if (sum === n) {
-                res.push([...path]);
-            }
-            return;
+    let res = [];
+    let path = [];
+    let sum = 0;
+    const dfs = (path,index) => {
+        // 剪枝操作
+        if (sum > n){
+            return 
         }
-        for (let i = start; i <= 9 - (k - l) + 1; i++) {
+        if (path.length == k) {
+            if(sum == n){
+                res.push([...path]);
+                return 
+            }
+        }
+        for (let i = index; i <= 9 - (k-path.length) + 1;i++) {
             path.push(i);
-            backtrack(i + 1);
-            path.pop();
+            sum = sum + i;
+            index += 1;
+            dfs(path,index);
+            sum -= i
+            path.pop()
         }
     }
-    let res = [], path = [];
-    backtrack(1);
-    return res;
+    dfs(path,1);
+    return res
 };
 ```