From 4aea70887c44a49b140d543eed9d7bfafa057d95 Mon Sep 17 00:00:00 2001
From: youngyangyang04 <826123027@qq.com>
Date: Wed, 23 Sep 2020 15:24:44 +0800
Subject: [PATCH] Update
---
README.md | 1 +
problems/0028.实现strStr().md | 5 +
problems/0112.路径总和.md | 81 ++++++++++++
problems/0459.重复的子字符串.md | 6 +
problems/0617.合并二叉树.md | 123 +++++++++++++++++-
problems/0968.监控二叉树.md | 12 +-
video/617.合并二叉树.gif | Bin 0 -> 1717578 bytes
.../二叉树中序遍历（迭代法）.gif | Bin 0 -> 2751512 bytes
.../二叉树中序遍历（迭代法）.mp4 | Bin 0 -> 180269 bytes
.../二叉树前序遍历（迭代法）.gif | Bin 0 -> 2389974 bytes
.../二叉树前序遍历（迭代法）.mp4 | Bin 0 -> 174871 bytes
11 files changed, 219 insertions(+), 9 deletions(-)
create mode 100644 problems/0112.路径总和.md
create mode 100644 video/617.合并二叉树.gif
create mode 100644 video/二叉树中序遍历（迭代法）.gif
create mode 100644 video/二叉树中序遍历（迭代法）.mp4
create mode 100644 video/二叉树前序遍历（迭代法）.gif
create mode 100644 video/二叉树前序遍历（迭代法）.mp4
diff --git a/README.md b/README.md
index a8427da5..9ac4f7ec 100644
--- a/README.md
+++ b/README.md
@@ -79,6 +79,7 @@
* [栈与队列：滑动窗口里求最大值引出一个重要数据结构](https://mp.weixin.qq.com/s/8c6l2bO74xyMjph09gQtpA)
* [栈与队列：求前 K 个高频元素和队列有啥关系？](https://mp.weixin.qq.com/s/8hMwxoE_BQRbzCc7CA8rng)
* [栈与队列：总结篇！](https://mp.weixin.qq.com/s/xBcHyvHlWq4P13fzxEtkPg)
+
* 二叉树
* [关于二叉树，你该了解这些！](https://mp.weixin.qq.com/s/_ymfWYvTNd2GvWvC5HOE4A)
diff --git a/problems/0028.实现strStr().md b/problems/0028.实现strStr().md
index 273582bb..3b151fb8 100644
--- a/problems/0028.实现strStr().md
+++ b/problems/0028.实现strStr().md
@@ -27,6 +27,11 @@ https://leetcode-cn.com/problems/implement-strstr/
本题是KMP 经典题目。
+以下文字如果看不进去，可以看我的B站视频：
+
+* [帮你把KMP算法学个通透！B站（理论篇）](https://www.bilibili.com/video/BV1PD4y1o7nd/)
+* [帮你把KMP算法学个通透！（求next数组代码篇）](https://www.bilibili.com/video/BV1M5411j7Xx)
+
KMP的经典思想就是:**当出现字符串不匹配时，可以记录一部分之前已经匹配的文本内容，利用这些信息避免从头再去做匹配。**
本篇将以如下顺序来讲解KMP，
diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md
new file mode 100644
index 00000000..9c8e6a87
--- /dev/null
+++ b/problems/0112.路径总和.md
@@ -0,0 +1,81 @@
+## 题目地址
+
+## 思路
+// 遍历单条边，还是遍历整个树，取最优数值！！
+// 对啊，用sum来减法啊，免得多定义一个变量
+
+## C++
+
+贼粗糙的写法
+
+深度优先遍历
+
+```
+class Solution {
+private:
+ bool traversal(TreeNode* cur, int count) {
+ if (!cur->left && !cur->right && count == 0) return true; // 遇到叶子节点，并且计数为0
+ if (!cur->left && !cur->right) return false; // 遇到叶子节点直接返回
+
+ if (cur->left) { // 左
+ // 遇到叶子节点返回true，则直接返回true
+ if (traversal(cur->left, count - cur->left->val)) return true;
+ }
+ if (cur->right) { // 右
+ // 遇到叶子节点返回true，则直接返回true
+ if (traversal(cur->right, count - cur->right->val)) return true;
+ }
+ return false;
+ }
+
+public:
+ bool hasPathSum(TreeNode* root, int sum) {
+ if (root == NULL) return false;
+ return traversal(root, sum - root->val);
+ }
+};
+```
+
+其实本题一定是有回溯的，没有回溯，如果后撤重新找另一条路径呢，但是貌似以上代码中，**大家貌似没有感受到回溯，那是因为回溯在代码里隐藏起来了。**
+
+隐藏在`traversal(cur->left, count - cur->left->val)`这里， 因为把`count - cur->left->val` 直接作为参数传进去，函数结束，count自然恢复到原先的数值了。
+
+为了把回溯的过程体现出来，将`if (traversal(cur->left, count - cur->left->val)) return true;` 改为如下代码：
+
+```
+if (cur->left) { // 左
+ count -= cur->left->val; // 递归，处理节点;
+ if (traversal(cur->left, count)) return true;
+ count += cur->left->val; // 回溯，撤销处理结果
+}
+```
+
+这样大家就能感受到回溯了，整体回溯代码如下：
+
+```
+class Solution {
+private:
+ bool traversal(TreeNode* cur, int count) {
+ if (!cur->left && !cur->right && count == 0) return true; // 遇到叶子节点，并且计数为0
+ if (!cur->left && !cur->right) return false; // 遇到叶子节点直接返回
+
+ if (cur->left) { // 左
+ count -= cur->left->val; // 递归，处理节点;
+ if (traversal(cur->left, count)) return true;
+ count += cur->left->val; // 回溯，撤销处理结果
+ }
+ if (cur->right) { // 右
+ count -= cur->right->val; // 递归，处理节点;
+ if (traversal(cur->right, count)) return true;
+ count += cur->right->val; // 回溯，撤销处理结果
+ }
+ return false;
+ }
+
+public:
+ bool hasPathSum(TreeNode* root, int sum) {
+ if (root == NULL) return false;
+ return traversal(root, sum - root->val);
+ }
+};
+```
diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md
index 6397c1fb..da163490 100644
--- a/problems/0459.重复的子字符串.md
+++ b/problems/0459.重复的子字符串.md
@@ -26,6 +26,12 @@ https://leetcode-cn.com/problems/repeated-substring-pattern/
这又是一道标准的KMP的题目。
+如果KMP还不够了解，可以看我的B站：
+
+* [帮你把KMP算法学个通透！B站（理论篇）](https://www.bilibili.com/video/BV1PD4y1o7nd/)
+* [帮你把KMP算法学个通透！（求next数组代码篇）](https://www.bilibili.com/video/BV1M5411j7Xx)
+
+
如果KMP还不够了解，可以看我的这个视频[帮你把KMP算法学个通透！B站](https://www.bilibili.com/video/BV1PD4y1o7nd/)
我们在[字符串：都来看看KMP的看家本领！](https://mp.weixin.qq.com/s/Gk9FKZ9_FSWLEkdGrkecyg)里提到了，在一个串中查找是否出现过另一个串，这是KMP的看家本领。
diff --git a/problems/0617.合并二叉树.md b/problems/0617.合并二叉树.md
index 43dcf67b..92eb7d11 100644
--- a/problems/0617.合并二叉树.md
+++ b/problems/0617.合并二叉树.md
@@ -3,7 +3,121 @@ https://leetcode-cn.com/problems/merge-two-binary-trees/
## 思路
-四种写法，总有一款适合你，其实这道题目迭代法实现是比较困难的，大家可以试一试，是一道不错的面试进阶题目。
+相信这道题目很多同学疑惑的点是如何同时遍历两个二叉树呢？
+
+其实和遍历一个树逻辑是一样的，只不过传入两个树的节点，同时操作。
+
+那么前中后序应该使用哪种遍历呢？
+
+**本题使用哪种遍历都是可以的！**
+
+我们下面以前序遍历为例。
+
+动画如下：
+
+
+
+
+那么我们来按照递归三部曲来解决：
+
+1. **确定递归函数的参数和返回值：**
+首先那么要合入两个二叉树，那么参数至少是要传入两个二叉树的根节点，返回值就是合并之后二叉树的根节点。
+
+代码如下：
+
+```
+TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
+```
+
+2. **确定终止条件：**
+
+因为是传入了两个树，那么就有两个树遍历的节点t1 和 t2，如果t1 == NULL 了，两个树合并就应该是 t2 了啊（如果t2也为NULL也无所谓）。
+
+反过来如果t2 == NULL，那么两个数合并就是t1（如果t1也为NULL也无所谓）。
+
+代码如下：
+
+```
+if (t1 == NULL) return t2; // 如果t1为空，合并之后就应该是t2
+if (t2 == NULL) return t1; // 如果t2为空，合并之后就应该是t1
+```
+
+
+3. **确定单层递归的逻辑：**
+
+单层递归的逻辑就比较好些了，这里我们用重复利用一下t1这个树，t1就是合并之后树的根节点（所谓的修改了元数据的结构）。
+
+那么单层递归中，就要把两棵树的元素加到一起。
+```
+t1->val += t2->val;
+```
+
+那么此时t1 的左子树 应该是 合并 t1左子树 t2左子树之后的左子树，t1 的右子树 应该是 合并 t1右子树 t2右子树之后的右子树。
+
+代码如下：
+
+```
+ t1->left = mergeTrees(t1->left, t2->left);
+ t1->right = mergeTrees(t1->right, t2->right);
+ return t1;
+```
+
+此时前序遍历，修改原输入树结构的完整代码就写出来了，如下：
+
+```
+class Solution {
+public:
+ TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
+ if (t1 == NULL) return t2; // 如果t1为空，合并之后就应该是t2
+ if (t2 == NULL) return t1; // 如果t2为空，合并之后就应该是t1
+ // 修改了t1的数值和结构
+ t1->val += t2->val; // 中
+ t1->left = mergeTrees(t1->left, t2->left); // 左
+ t1->right = mergeTrees(t1->right, t2->right); // 右
+ return t1;
+ }
+};
+```
+
+那么中序遍历可不可以呢，也是可以的，代码如下：
+
+```
+class Solution {
+public:
+ TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
+ if (t1 == NULL) return t2; // 如果t1为空，合并之后就应该是t2
+ if (t2 == NULL) return t1; // 如果t2为空，合并之后就应该是t1
+ // 修改了t1的数值和结构
+ t1->left = mergeTrees(t1->left, t2->left); // 左
+ t1->val += t2->val; // 中
+ t1->right = mergeTrees(t1->right, t2->right); // 右
+ return t1;
+ }
+};
+```
+
+后序遍历呢，依然可以，代码如下：
+
+```
+class Solution {
+public:
+ TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
+ if (t1 == NULL) return t2; // 如果t1为空，合并之后就应该是t2
+ if (t2 == NULL) return t1; // 如果t2为空，合并之后就应该是t1
+ // 修改了t1的数值和结构
+ t1->left = mergeTrees(t1->left, t2->left); // 左
+ t1->right = mergeTrees(t1->right, t2->right); // 右
+ t1->val += t2->val; // 中
+ return t1;
+ }
+};
+```
+
+**但是前序遍历是最好理解的，我建议大家用前序遍历来做就OK。**
+
+**那么如下还总结了四种方法，递归的方式均使用了前序遍历，此时大家应该知道了，以下每一种递归的方法都可以换成中序和后序遍历，所以本题的解法是很多的。**
+
+**其实这道题目迭代法实现是比较困难的，大家可以试一试，是一道不错的面试进阶题目。**
四种写法如下：
@@ -16,7 +130,7 @@ https://leetcode-cn.com/problems/merge-two-binary-trees/
### 递归
-修改了输入树的结构
+修改了输入树的结构，前序遍历
```
class Solution {
public:
@@ -32,7 +146,7 @@ public:
};
```
-不修改输入树的结构
+不修改输入树的结构，前序遍历
```
class Solution {
public:
@@ -51,6 +165,7 @@ public:
一波指针的操作，自己写的野路子
想要更改二叉树的值，应该传入指向指针的指针， 如果process(t1, t2);这么写的话，其实只是传入的一个int型的指针，并没有传入地址，要传入指向指针的指针才能完成对t1的修改。
+（前序遍历）
```
class Solution {
public:
@@ -77,7 +192,7 @@ public:
```
### 迭代
-这应该是最简单直观的迭代法了
+这应该是最简单直观的迭代法了，模拟的层序遍历。
```
class Solution {
public:
diff --git a/problems/0968.监控二叉树.md b/problems/0968.监控二叉树.md
index 613a5a5c..dc3fec03 100644
--- a/problems/0968.监控二叉树.md
+++ b/problems/0968.监控二叉树.md
@@ -12,13 +12,13 @@ https://leetcode-cn.com/problems/binary-tree-cameras/
我们之前做动态规划的时候，只要最难的地方在于确定状态转移方程，至于遍历方式无非就是在数组或者二维数组上。
-**而本题，不仅要确定状态转移方式，而且要在树上进行推导，所以难度就上来了，一些同学知道这道题目难，但其实说不上难点究竟在哪。**
+**本题并不是动态规划，其本质是贪心，但我们要确定状态转移方式，而且要在树上进行推导，所以难度就上来了，一些同学知道这道题目难，但其实说不上难点究竟在哪。**
1. 需要确定遍历方式
首先先确定遍历方式，才能确定转移方程，那么该如何遍历呢？
-在安排选择摄像头的位置的时候，**我们要从底向上进行推导，因为尽量让叶子节点的父节点安装摄像头，这样摄像头的数量才是最少的**
+在安排选择摄像头的位置的时候，**我们要从底向上进行推导，因为尽量让叶子节点的父节点安装摄像头，这样摄像头的数量才是最少的**，这也是本道贪心的原理所在！
如何从低向上推导呢？
@@ -59,9 +59,11 @@ https://leetcode-cn.com/problems/binary-tree-cameras/
* 1：本节点有摄像头
* 2：本节点有覆盖
-大家应该找不出第四个节点的状态了。
+大家应该找不出第四个节点的状态了。
-那么问题来了，空节点究竟是哪一种状态呢？ 空节点表示无覆盖？ 表示有摄像头？还是有覆盖呢？
+**一些同学可能会想有没有第四种状态：本节点无摄像头，其实无摄像头就是 无覆盖 或者 有覆盖的状态，所以一共还是三个状态。**
+
+**那么问题来了，空节点究竟是哪一种状态呢？ 空节点表示无覆盖？ 表示有摄像头？还是有覆盖呢？ **
回归本质，为了让摄像头数量最少，我们要尽量让叶子节点的父节点安装摄像头，这样才能摄像头的数量最少。
@@ -141,7 +143,7 @@ left == 1 && right == 1 左右节点都有摄像头
这种情况也是大多数同学容易迷惑的情况。
-4. 情况4
+4. 情况4：头结点没有覆盖
以上都处理完了，递归结束之后，可能头结点 还有一个无覆盖的情况，如图：
diff --git a/video/617.合并二叉树.gif b/video/617.合并二叉树.gif
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