404. Sum of Left Leaves [题目]中文版  英文版 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int sumOfLeftLea…

404. Sum of Left Leaves Easy Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. package leetcode.easy; /** * Definition for…

404. 左叶子之和 404. Sum of Left Leaves LeetCode404. Sum of Left Leaves 题目描述 计算给定二叉树的所有左叶子之和. 示例: 3 / \ 9 20 / \ 15 7 在这个二叉树中,有两个左叶子,分别是 9 和 15,所以返回 24. Java 实现 TreeNode 结构 class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x…

Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. 这道题让我们求一棵二叉树的所有左子叶的和,那么看到这道题我们知道这肯定是考二叉树的遍历问题,那么最简洁的写法肯定是用递归,由于我们只需要累加左子叶…

Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. /** * Definition for a binary tree node. * public class TreeNode { * int…

原题链接在这里:https://leetcode.com/problems/sum-of-left-leaves/ 题目: Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. 题解: DFS, 若r…

Find the sum of all left leaves in a given binary tree. 左树的值(9+15=24) /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { publi…

Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. /** * Definition for a binary tree node. * public class TreeNode { * int…

Find the sum of all left leaves in a given binary tree. Example:     3    / \   9  20     /  \    15   7   There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. 思路:还是递归.仔细看代码,体会递归的设计方法.…

Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. 题目标签:Tree 这道题目给了我们一个二叉树,让我们找到所有左子叶之和.这里需要另外一个function - sumLeftLeaves 还要一…

Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. 计算给定二叉树的所有左叶子之和. 示例: 3 / \ 9 20 / \ 15 7 在这个二叉树中,有两个左叶子,分别是 9 和 15,所以返回 2…

Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. # Definition for a binary tree node. # class TreeNode(object): # def __in…

题目: Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. 分析: 给定一颗二叉树,求左叶子节点的和. 重点在于如何判断左叶子节点,如果一个节点的left存在,且left的left和right都为空…

［抄题］: Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. [暴力解法]: 时间分析: 空间分析: ［奇葩输出条件］: ［奇葩corner case］: ［思维问题］: root.left ro…

Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24. Solution 1:BFS /** * Definition for a binary tree node. * public class Tr…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/4084 访问. 计算给定二叉树的所有左叶子之和. 3      / \    9  20   /       \ 15       7 在这个二叉树中,有两个左叶子,分别是 9 和 15,所以返回 24 Find the sum of all left leaves in a given binary tree. 3      / \    9  20   /  …

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目大意 题目大意 解题方法 递归 迭代 日期 [LeetCode] 题目地址:https://leetcode.com/problems/sum-of-left-leaves/ Difficulty: Easy 题目大意 Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \…

这是悦乐书的第217次更新,第230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85题(顺位题号是404).找到给定二叉树中所有左叶的总和.例如: 二叉树中有两个左叶,分别为9和15. 返回24. 3 / \ 9 20 / \ 15 7 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试. 02 第一种解法 使用递归. 特殊情况:当root为null时,直接返回0. 正常情况:定义一个变量su…

------------------------------------------------------------------- 分两种情况: 1.当前节点拥有左孩子并且左孩子是叶子节点:左孩子值+右孩子子树遍历统计2.不符合上面那种情况的从当前节点劈开为两颗子树分别统计相加即可 AC代码: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * Tre…

#-*- coding: UTF-8 -*- # Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):        def dfs(self,root,isLeft):      …

求所有左节点的和. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int sumOfLeftLeaves(TreeNode* root) { if(ro…

1.题目描述 2.问题分析 对于每个节点,如果其左子节点是叶子,则加上它的值,如果不是,递归,再对右子节点递归即可. 3.代码 int sumOfLeftLeaves(TreeNode* root) { if (root == NULL) ; ; if (root->left != NULL) { if (root->left->left == NULL && root->left->right == NULL) ans += root->left-&g…

弄个flag记录是不是左节点就行 int res = 0; public int sumOfLeftLeaves(TreeNode root) { if (root==null) return res; leftSum(root,false); return res; } private void leftSum(TreeNode root,boolean flag) { if (root.left==null&&root.right==null&&flag) res+=r…

今天看到这道题目:http://www.cnblogs.com/charlesblc/p/5930311.html 题目地址:https://leetcode.com/problems/split-array-largest-sum/ 很好,也很难.开拓了思路,用二分法来查找结果备选,然后直接划分原集合,来反过来是否是合理的解.是或者不是的话,继续向相应的方向进行二分.   # Title Editorial Acceptance Difficulty Frequency   . 454 4Su…

要求 给出一个二叉树及数字sum,判断是否存在一条从根到叶子的路径,路径上的所有节点和为sum 实现 转化为寻找左右子树上和为 sum-root 的路径,到达叶子节点时递归终止 注意只有一个孩子时,根节点本身不构成一条路径,如下图sum=5的情况,终止条件是不对的 1 class Solution { 2 public: 3 bool hasPathSum(TreeNode* root, int sum) { 4 5 if( root == NULL ) 6 return false; 7 8…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance 44.10% Meidum 475 Heaters  30.20% Easy 474 Ones and Zeroes  34.90% Meidum 473 Matchsticks to Square  31.80% Medium 472 Concatenated Words 29.20% Hard…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be its length. Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow: F(k…

Python不熟悉 不同的做法 404. Sum of Left Leaves 这是我的做法,AC. class Solution(object): res = 0 def recursive(self, root): if root == None: return if root.left != None and root.left.left == None and root.left.right == None: self.res += root.left.val self.recursiv…

参考资料: <复杂数据统计方法>&网络&帮助文件 适用情况:在因变量为分类变量而自变量含有多个分类变量或分类变量水平较多的情况. 一. (一)概论和例子 数据来源:http://archive.ics.uci.edu/ml/datasets/Cardiotocography 自变量:LB - FHR baseline (beats per minute) AC - # of accelerations per secondFM - # of fetal movements pe…