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update: 添加问题“1022.从根到叶的二进制数之和”的代码(并更新其题解) (#1409)
1022: AC.cpp (#1408) - AC,100.00%,78.62% + en(@beijing) Signed-off-by: LetMeFly666 <Tisfy@qq.com>
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‎.commitmsg‎

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1461: AC.cpp (#1406) - AC,40.08%,36.38%
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1022: AC.cpp (#1408) - AC,100.00%,78.62% + en(@Beijing)

‎Codes/1022-sum-of-root-to-leaf-binary-numbers_20260224.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2026-02-24 21:34:35
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2026-02-24 21:37:38
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*/
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#if defined(_WIN32) || defined(__APPLE__)
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#include "_[1,2]toVector.h"
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#endif
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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private:
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int ans;
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// @input.root not empty
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void dfs(TreeNode* root, int now) {
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now <<= 1;
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now += root->val;
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if (!root->left && !root->right) {
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ans += now;
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return;
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}
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if (root->left) {
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dfs(root->left, now);
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}
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if (root->right) {
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dfs(root->right, now);
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}
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}
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public:
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int sumRootToLeaf(TreeNode* root) {
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ans = 0;
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dfs(root, 0);
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return ans;
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}
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};

‎Codes/1461-check-if-a-string-contains-all-binary-codes-of-size-k.cpp‎

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* @Author: LetMeFly
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* @Date: 2026-02-23 11:30:05
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2026-02-23 11:37:00
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* @LastEditTime: 2026-02-23 11:37:10
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*/
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#if defined(_WIN32) || defined(__APPLE__)
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#include "_[1,2]toVector.h"
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#endif
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// THIS CANNOT BE ACCEPTED
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class Solution {
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public:
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bool hasAllCodes(string s, int k) {
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unordered_set<string> ma;
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for (int i = 0, n = s.size(); i < n; i++) {
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for (int l = 1; l <= k && i + l <= n; l++) {
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ma.insert(s.substr(i, l));
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}
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for (int i = 0, to = s.size() - k + 1; i < to; i++) {
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ma.insert(s.substr(i, k));
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}
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return ma.size() == (1 << k);
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}

‎Codes/1461-check-if-a-string-contains-all-binary-codes-of-size-k_AC.cpp‎

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This file was deleted.

‎Solutions/LeetCode 1022.从根到叶的二进制数之和.md‎

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---
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title: 1022.从根到叶的二进制数之和
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title: 1022.从根到叶的二进制数之和：DFS或BFS
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date: 2022-05-30 10:52:40
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tags: [题解, LeetCode, 简单, 二叉树, 树, 交互, 模拟]
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tags: [题解, LeetCode, 简单, 二叉树, 树, 模拟, 深度优先搜索, DFS, 广度优先搜索, BFS, 递归, 位运算]
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categories: [题解, LeetCode]
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index_img: https://assets.leetcode.com/uploads/2019/04/04/sum-of-root-to-leaf-binary-numbers.png
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---
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# 【LetMeFly】1022.从根到叶的二进制数之和
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# 【LetMeFly】1022.从根到叶的二进制数之和：DFS或BFS
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力扣题目链接：[https://leetcode.cn/problems/sum-of-root-to-leaf-binary-numbers/](https://leetcode.cn/problems/sum-of-root-to-leaf-binary-numbers/)
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二进制的100等于十进制的4
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# 思路
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我们只需要层次遍历这棵树，遍历到某个节点时，如果存在子节点，子节点就加上这个节点的“值的<<1”的结果
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# 解题方法
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## 方法一：层次遍历
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我们只需要层次遍历这棵树，遍历到某个节点时，如果存在子节点，子节点就加上这个节点的“值的<<1”的结果
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+ 时间复杂度$O(n)$，其中$n$是树中节点的数量
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+ 空间复杂度$O(n)$
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#### C++
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```cpp
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/*
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* @LastEditTime: 2022-05-30 10:50:08
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*/
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class Solution {
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public:
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int sumRootToLeaf(TreeNode* root) { // 不会为空
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};
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```
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> 同步发文于CSDN，原创不易，转载请附上[原文链接](https://blog.letmefly.xyz/2022/05/30/LeetCode%201022.%E4%BB%8E%E6%A0%B9%E5%88%B0%E5%8F%B6%E7%9A%84%E4%BA%8C%E8%BF%9B%E5%88%B6%E6%95%B0%E4%B9%8B%E5%92%8C)哦~
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> Tisfy：[https://letmefly.blog.csdn.net/article/details/125043202](https://letmefly.blog.csdn.net/article/details/125043202)
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## 方法二：深度优先搜索DFS
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使用一个类中的“全局变量”$ans$记录所有路径数字之和，写一个DFS函数：
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```cpp
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void dfs(TreeNode* root, int now)
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```
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其中`now`代表该走节点`root`时已经走过路径的二进制值。
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首先更新走到当前节点后的路径值`now = now << 1 + root->val`，如果`root`为叶节点则将该路径值累加到$ans$，否则继续递归左右子中非空的节点。
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+ 时间复杂度$O(n)$，其中$n$是树中节点的数量
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+ 空间复杂度$O(n)$
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### AC代码
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#### C++
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```cpp
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/*
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* @LastEditTime: 2026-02-24 21:37:38
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*/
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class Solution {
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private:
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int ans;
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// @input.root not empty
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void dfs(TreeNode* root, int now) {
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now <<= 1;
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now += root->val;
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if (!root->left && !root->right) {
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ans += now;
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return;
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}
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if (root->left) {
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dfs(root->left, now);
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}
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if (root->right) {
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dfs(root->right, now);
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}
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}
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public:
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int sumRootToLeaf(TreeNode* root) {
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ans = 0;
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dfs(root, 0);
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return ans;
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}
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};
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```
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> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/125043202)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](ttps://blog.letmefly.xyz/2022/05/30/LeetCode%201022.%E4%BB%8E%E6%A0%B9%E5%88%B0%E5%8F%B6%E7%9A%84%E4%BA%8C%E8%BF%9B%E5%88%B6%E6%95%B0%E4%B9%8B%E5%92%8C)哦~
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>
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> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)

‎Solutions/Other-English-LearningNotes-SomeWords.md‎

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|stipulation|n. 规定，约定，合同|
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|turnip|n. 萝卜|
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|senseless|adj. 无意义的，无目的的；失去知觉的；不明智的，愚蠢的|
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|enlightening|v. 启迪，指导<br/>adj. 有启迪的，使人感悟的|
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|potent|adj. 强有效的，有力的，烈性的，影响身心的|
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+ 这个web要是能设计得可以闭眼(完全不睁眼)键盘控制背单词就好了。
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+ 也许可以加个AI用最近词编故事功能(返回接口中支持标注所使用单词高亮?)
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</script>
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> 原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2024/02/19/Other-English-LearningNotes-SomeWords/)哦~
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> [https://blog.letmefly.xyz/2024/02/19/Other-English-LearningNotes-SomeWords/](https://blog.letmefly.xyz/2024/02/19/Other-English-LearningNotes-SomeWords/)
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> [https://blog.letmefly.xyz/2024/02/19/Other-English-LearningNotes-SomeWords/](https://blog.letmefly.xyz/2024/02/19/Other-English-LearningNotes-SomeWords/)

‎todo‎

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# rm Codes\1461-check-if-a-string-contains-all-binary-codes-of-size-k_AC.cpp
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# 记录iphone快捷指令 自动化 当到达某位置时触发
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# move beautify.sh to github repo/gist
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else:
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return getproxies()
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print(get_environ_proxies("https://letmefly.xyz"))
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print(get_environ_proxies("https://letmefly.xyz"))

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