更新了113和113的说明

Author: LuYanFCP

README.md

@@ -2,3 +2,10 @@
 
 此为我的Leetcode刷题记录
 
+## 分类整理
+
+### 二叉树
+
+#### 二叉树的遍历
+
++ No.113: [cpp](cpp/113.cpp) [python](python/113.py)

cpp/113.cpp

@@ -0,0 +1,49 @@
+#include <vector>
+#include <iostream>
+using std::vector;
+
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+};
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+
+/* 16 ms 69.96%	
+ * 20 MB 50.00%
+ */
+class Solution {
+public:
+    vector<vector<int>> res;
+    int sum;
+    vector<vector<int>> pathSum(TreeNode* root, int sum) {
+        vector<int> stack;
+        this->sum = sum;
+        order(root, stack, 0);
+        return res;
+    }
+    void order(TreeNode* root, vector<int>& stack,int sum) {
+        if (root != nullptr) {
+            stack.push_back(root->val);
+            sum += root->val;
+            if (root->left || root->right) {  // 是否是根节点
+                order(root->left, stack, sum);
+                order(root->right, stack, sum);
+            } else {
+                if (sum == this->sum) {
+                    this->res.push_back(stack);  // push_back为拷贝操作
+                }
+            }
+            stack.pop_back();
+        }
+    }
+};

python/113.py

@@ -0,0 +1,42 @@
+class Solution:        
+    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
+        self.res = []
+        self.target = sum
+        self.order(root, [], 0)
+        return self.res
+    
+    def order(self, root, ls, sum):
+        """
+        52 ms	
+        17.9 MB	
+        """
+        if root is not None:
+            ls += [root.val]
+            sum += root.val
+            if root.left != None or root.right != None:  # 叶子节点判断
+                self.inorder(root.left, ls[:], sum)  # ls[:]很浪费空间和时间, 因此order2的时候改成复用ls,加一个pop操作即可
+                self.inorder(root.right, ls[:], sum)
+            else:
+                if sum == self.target:
+                    # 这里会重复append
+                    self.res.append(ls)
+
+    def order(self, root, ls, sum):
+        """
+        44 ms	 72.95% 
+        13.8 MB  100%
+        """
+        if root is not None:
+            ls += [root.val]
+            sum += root.val
+            if root.left != None or root.right != None: 
+                self.inorder(root.left, ls, sum)  # 此处做了修改
+                self.inorder(root.right, ls, sum)
+            else:
+                if sum == self.target:
+                    # 这里会重复append
+                    self.res.append(ls[:])  # 这里还是copy操作,因为ls是引用
+            ls.pop() # 弹出栈
+            
+
+