From 9eea30cb9f4e9811aaf3aa8f087ac640fc703727 Mon Sep 17 00:00:00 2001
From: youngyangyang04 <826123027@qq.com>
Date: Tue, 6 Oct 2020 10:38:33 +0800
Subject: [PATCH] Update

---
 README.md                                  |   1 +
 pics/404.左叶子之和1.png              | Bin 0 -> 22375 bytes
 problems/0100.相同的树.md              |  30 +++---
 problems/0257.二叉树的所有路径.md  |   4 +
 problems/0404.左叶子之和.md           | 114 +++++++++++++++++----
 problems/二叉树中递归带着回溯.md |  72 +++++++++++++
 6 files changed, 185 insertions(+), 36 deletions(-)
 create mode 100644 pics/404.左叶子之和1.png
 create mode 100644 problems/二叉树中递归带着回溯.md

diff --git a/README.md b/README.md
index 3f432a75..18d36d58 100644
--- a/README.md
+++ b/README.md
@@ -94,6 +94,7 @@
     * [二叉树：我有多少个节点？](https://mp.weixin.qq.com/s/2_eAjzw-D0va9y4RJgSmXw)
     * [二叉树：我平衡么？](https://mp.weixin.qq.com/s/isUS-0HDYknmC0Rr4R8mww)
     * [二叉树：找我的所有路径？](https://mp.weixin.qq.com/s/Osw4LQD2xVUnCJ-9jrYxJA)
+    * [还在玩耍的你，该总结啦！（本周小结之二叉树）](https://mp.weixin.qq.com/s/QMBUTYnoaNfsVHlUADEzKg)
  
 
 （持续更新中....）
diff --git a/pics/404.左叶子之和1.png b/pics/404.左叶子之和1.png
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diff --git a/problems/0100.相同的树.md b/problems/0100.相同的树.md
index 6a771117..b596010a 100644
--- a/problems/0100.相同的树.md
+++ b/problems/0100.相同的树.md
@@ -68,40 +68,38 @@ else if (tree1->val != tree2->val) return false; // 注意这里我没有
 代码如下：
 
 ```
-bool outside = compare(tree1->left, tree2->right);   // 左子树：左、 右子树：左
-bool inside = compare(tree1->right, tree2->left);    // 左子树：右、 右子树：右
+bool outside = compare(tree1->left, tree2->left);   // 左子树：左、 右子树：左
+bool inside = compare(tree1->right, tree2->right);    // 左子树：右、 右子树：右
 bool isSame = outside && inside;                     // 左子树：中、 右子树：中（逻辑处理）
 return isSame;
 ```
--------------------------------------------------------------- 写到这
 最后递归的C++整体代码如下：
 
 ```
 class Solution {
 public:
-    bool compare(TreeNode* left, TreeNode* right) {
-        // 首先排除空节点的情况
-        if (left == NULL && right != NULL) return false;
-        else if (left != NULL && right == NULL) return false;
-        else if (left == NULL && right == NULL) return true;
-        // 排除了空节点，再排除数值不相同的情况
-        else if (left->val != right->val) return false;
+    bool compare(TreeNode* tree1, TreeNode* tree2) {
+        if (tree1 == NULL && tree2 != NULL) return false;
+        else if (tree1 != NULL && tree2 == NULL) return false;
+        else if (tree1 == NULL && tree2 == NULL) return true; 
+        else if (tree1->val != tree2->val) return false; // 注意这里我没有使用else
 
         // 此时就是：左右节点都不为空，且数值相同的情况
         // 此时才做递归，做下一层的判断
-        bool outside = compare(left->left, right->right);   // 左子树：左、 右子树：左
-        bool inside = compare(left->right, right->left);    // 左子树：右、 右子树：右
-        bool isSame = outside && inside;                    // 左子树：中、 右子树：中 （逻辑处理）
+        bool outside = compare(tree1->left, tree2->left);   // 左子树：左、 右子树：左
+        bool inside = compare(tree1->right, tree2->right);  // 左子树：右、 右子树：右
+        bool isSame = outside && inside;                    // 左子树：中、 右子树：中（逻辑处理）
         return isSame;
 
     }
-    bool isSymmetric(TreeNode* root) {
-        if (root == NULL) return true;
-        return compare(root->left, root->right);
+    bool isSameTree(TreeNode* p, TreeNode* q) {
+        return compare(p, q);
     }
 };
 ```
 
+-------------------------------------------------------------- 写到这
+
 **我给出的代码并不简洁，但是把每一步判断的逻辑都清楚的描绘出来了。**
 
 如果上来就看网上各种简洁的代码，看起来真的很简单，但是很多逻辑都掩盖掉了，而题解可能也没有把掩盖掉的逻辑说清楚。
diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md
index 3fcbe67d..1e083c78 100644
--- a/problems/0257.二叉树的所有路径.md
+++ b/problems/0257.二叉树的所有路径.md
@@ -207,8 +207,12 @@ public:
 
 那么在如上代码中，**貌似没有看到回溯的逻辑，其实不然，回溯就隐藏在`traversal(cur->left, path + "->", result);`中的 `path + "->"`。**  每次函数调用完，path依然是没有加上"->" 的，这就是回溯了。
 
+
+
 **综合以上，第二种递归的代码虽然精简但把很多重要的点隐藏在了代码细节里，第一种递归写法虽然代码多一些，但是把每一个逻辑处理都完整的展现了出来了。** 
 
+
+
 ## 迭代法 
 
 至于非递归的方式，我们可以依然可以使用前序遍历的迭代方式来模拟遍历路径的过程，对该迭代方式不了解的同学，可以看文章[二叉树：听说递归能做的，栈也能做！](https://mp.weixin.qq.com/s/c_zCrGHIVlBjUH_hJtghCg)和[二叉树：前中后序迭代方式的写法就不能统一一下么？](https://mp.weixin.qq.com/s/WKg0Ty1_3SZkztpHubZPRg)。
diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md
index d1a17f7b..4d8d5f04 100644
--- a/problems/0404.左叶子之和.md
+++ b/problems/0404.左叶子之和.md
@@ -1,9 +1,19 @@
 ## 题目地址 
 https://leetcode-cn.com/problems/sum-of-left-leaves/
 
-## 思路 
+> 有的题目就是
 
-首先要注意是判断左叶子呢，不是左子树，不要上来想着层序遍历。
+# 404.左叶子之和 
+
+计算给定二叉树的所有左叶子之和。
+
+示例：
+
+<img src='../pics/404.左叶子之和1.png' width=600> </img></div>
+
+# 思路 
+
+**首先要注意是判断左叶子，不是二叉树左侧节点，所以不要上来想着层序遍历。**
 
 因为题目中其实没有说清楚左叶子究竟是什么节点，那么我来给出左叶子的明确定义：**如果左节点不为空，且左节点没有左右孩子，那么这个节点就是左叶子**
 
@@ -13,42 +23,99 @@ https://leetcode-cn.com/problems/sum-of-left-leaves/
 
 **其实是0，因为这棵树根本没有左叶子！**
 
-那么判断左叶子，如果**判断当前节点是不是左叶子是无法判断的，必须要通过节点的父节点来判断其左孩子是不是左叶子。**
+那么**判断当前节点是不是左叶子是无法判断的，必须要通过节点的父节点来判断其左孩子是不是左叶子。**
 
-判断的根据如下：
+
+如果该节点的左节点不为空，该节点的左节点的左节点为空，该节点的左节点的右节点为空，则找到了一个左叶子，判断代码如下：
 
 ```
-    if (node->left != NULL && node->left->left == NULL && node->left->right == NULL) {
-        sum = node->left->val;
-    }
+if (node->left != NULL && node->left->left == NULL && node->left->right == NULL) {
+    左叶子节点处理逻辑
+}
 ```
 
-递归写法代码如下：
+## 递归法
 
-## 递归C++代码
+递归的遍历顺序为后序遍历（左右中），是因为要通过递归函数的返回值来累加求取左叶子数值之和。。
+
+递归三部曲：
+
+1. 确定递归函数的参数和返回值 
+
+判断一个树的左叶子节点之和，那么一定要传入树的根节点，递归函数的返回值为数值之和，所以为int 
+
+使用题目中给出的函数就可以了。
+
+2. 确定终止条件 
+
+依然是
+```
+if (root == NULL) return 0;
+```
+
+3. 确定单层递归的逻辑  
+
+当遇到左叶子节点的时候，记录数值，然后通过递归求取左子树左叶子之和，和 右子树左叶子之和，相加便是整个树的左叶子之和。
+
+代码如下：
+
+```
+int leftValue = sumOfLeftLeaves(root->left);    // 左
+int rightValue = sumOfLeftLeaves(root->right);  // 右
+                                                // 中
+int midValue = 0;
+if (root->left && !root->left->left && !root->left->right) { 
+    midValue = root->left->val;
+}
+int sum = midValue + leftValue + rightValue;
+return sum;
+
+```
+
+
+整体递归代码如下：
 
 ```
 class Solution {
 public:
     int sumOfLeftLeaves(TreeNode* root) {
         if (root == NULL) return 0;
-        int sum = 0;
-        if (root->left != NULL && root->left->left == NULL && root->left->right == NULL) {
-            sum = root->left->val;
+
+        int leftValue = sumOfLeftLeaves(root->left);    // 左
+        int rightValue = sumOfLeftLeaves(root->right);  // 右
+                                                        // 中
+        int midValue = 0;
+        if (root->left && !root->left->left && !root->left->right) { // 中
+            midValue = root->left->val;
         }
-        return sum + sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
+        int sum = midValue + leftValue + rightValue;
+        return sum;
     }
 };
 ```
 
-递归的过程其实就是二叉树的前序遍历，那么写过二叉树的同学都知道，既然是二叉树的前序遍历，能写出递归，就能写出非递归。
+以上代码精简之后如下：
 
-如果对二叉树的各种递归和非递归的写法不熟悉，可以看我的这个题解：
-[彻底吃透二叉树的前中后序递归法和迭代法！！](https://leetcode-cn.com/problems/binary-tree-inorder-traversal/solution/che-di-chi-tou-er-cha-shu-de-qian-zhong-hou-xu-d-2/)
+```
+class Solution {
+public:
+    int sumOfLeftLeaves(TreeNode* root) {
+        if (root == NULL) return 0;
+        int midValue = 0;
+        if (root->left != NULL && root->left->left == NULL && root->left->right == NULL) {
+            midValue = root->left->val;
+        }
+        return midValue + sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
+    }
+};
+```
 
-那么非递归版本善良登场，判断条件都是一样的
+## 迭代法 
 
-## 非递归C++代码
+
+本题使用了后序遍历啊，那么参考文章 [二叉树：听说递归能做的，栈也能做！](https://mp.weixin.qq.com/s/c_zCrGHIVlBjUH_hJtghCg)和[二叉树：前中后序迭代方式的写法就不能统一一下么？](https://mp.weixin.qq.com/s/WKg0Ty1_3SZkztpHubZPRg)中的写法，同样可以写出一个后序遍历的迭代法。
+
+判断条件都是一样的，代码如下：
 
 ```
 
@@ -70,8 +137,15 @@ public:
         }
         return result;
     }
-
-
 };
 ```
 
+# 总结 
+
+这道题目要求左叶子之和，其实是比较绕的，因为不能判断本节点是不是左叶子节点。
+
+此时就要通过节点的父节点来判断其左孩子是不是左叶子了。 
+
+**平时我们解二叉树的题目时，已经习惯了通过节点的左右孩子判断本节点的属性，而本题我们要通过节点的父节点判断本节点的属性。**
+
+希望通过这道题目，可以扩展大家对二叉树的解题思路。
diff --git a/problems/二叉树中递归带着回溯.md b/problems/二叉树中递归带着回溯.md
new file mode 100644
index 00000000..3cb00f24
--- /dev/null
+++ b/problems/二叉树中递归带着回溯.md
@@ -0,0 +1,72 @@
+
+在上一面 
+
+
+```
+class Solution {
+private:
+    void traversal(TreeNode* cur, string path, vector<string>& result) {
+        path += to_string(cur->val); // 中
+        if (cur->left == NULL && cur->right == NULL) {
+            result.push_back(path);
+            return;
+        }
+
+        if (cur->left) {
+            path += "->";
+            traversal(cur->left, path, result); // 左
+            path.pop_back(); // 回溯
+            path.pop_back(); 
+        }
+        if (cur->right) {
+            path += "->";
+            traversal(cur->right, path, result); // 右
+            path.pop_back(); // 回溯
+            path.pop_back();
+        }
+    }
+
+public:
+    vector<string> binaryTreePaths(TreeNode* root) {
+        vector<string> result;
+        string path;
+        if (root == NULL) return result;
+        traversal(root, path, result);
+        return result;
+
+    }
+};
+```
+
+没有回溯了
+```
+class Solution {
+private:
+    void traversal(TreeNode* cur, string path, vector<string>& result) {
+        path += to_string(cur->val); // 中
+        if (cur->left == NULL && cur->right == NULL) {
+            result.push_back(path);
+            return;
+        }
+
+        if (cur->left) {
+            path += "->";
+            traversal(cur->left, path, result); // 左
+        }
+        if (cur->right) {
+            path += "->";
+            traversal(cur->right, path, result); // 右
+        }
+    }
+
+public:
+    vector<string> binaryTreePaths(TreeNode* root) {
+        vector<string> result;
+        string path;
+        if (root == NULL) return result;
+        traversal(root, path, result);
+        return result;
+
+    }
+};
+```