problem 1022. Sum of Root To Leaf Binary Numbers 参考 1. Leetcode_easy_1022. Sum of Root To Leaf Binary Numbers; 完…

题目如下: Given a binary tree, each node has value 0 or 1.  Each root-to-leaf path represents a binary number starting with the most significant bit.  For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, wh…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS 日期 题目地址:https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/ 题目描述 Given a binary tree, each node has value 0 or 1. Each root-to-leaf path represents a binary n…

网址:https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/ 递归调用求和,同时注意%1000000007的位置 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(…

dfs class Solution: def sumRootToLeaf(self, root: TreeNode) -> int: stack=[(root,0)] ans=[] bi_str=[] while stack: node,level=stack.pop() s_len=len(bi_str) if s_len-1<level: bi_str.append(str(node.val)) else: bi_str[level]=str(node.val) del bi_str[l…

1022. 从根到叶的二进制数之和 1022. Sum of Root To Leaf Binary Numbers 题目描述 Given a binary tree, each node has value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit. For example, if the path is 0 -> 1 -> 1 ->…

这是小川的第381次更新,第410篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第243题(顺位题号是1022).给定二叉树,每个节点值为0或1.每个根到叶路径表示以最高有效位开始的二进制数.例如,如果路径为0 -> 1 -> 1 -> 0 -> 1,那么这可能表示二进制的01101,即13. 对于树中的所有叶子节点,请考虑从根到该叶子节点的路径所代表的数字.返回这些数字的总和. 例如: 1 / \ 0 1 / \ / \ 0 1 0 1 输入:[1,…

Given a binary tree, each node has value 0 or 1.  Each root-to-leaf path represents a binary number starting with the most significant bit.  For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is…

problem 985. Sum of Even Numbers After Queries class Solution { public: vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) { vector<int> res; ; ==) sum +=a; for(auto query:queries) { ]]%== )…

problem 633. Sum of Square Numbers 题意: solution1: 可以从c的平方根,注意即使c不是平方数,也会返回一个整型数.然后我们判断如果 i*i 等于c,说明c就是个平方数,只要再凑个0,就是两个平方数之和,返回 true:如果不等于的话,那么算出差值 c - i*i,如果这个差值也是平方数的话,返回 true.遍历结束后返回 false, class Solution { public: bool judgeSquareSum(int c) { ; --…