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…

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…

［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. 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…

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…

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.…

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…

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 class Solution { public: int sum = 0; int sumOfLeftLeaves(TreeNode* root) { if(root == NULL) return 0; if(root ->left == NULL && root ->right == NULL) return 0;…

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 / \…

------------------------------------------------------------------- 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…

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…

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. Solution 1:BFS /** * Definition for a binary tree node. * public class Tr…

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…

We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not…

Given an array of integers A, consider all non-empty subsequences of A. For any sequence S, let the width of S be the difference between the maximum and minimum element of S. Return the sum of the widths of all subsequences of A.  As the answer may b…

Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c. Example 1: Input: 5 Output: True Explanation: 1 * 1 + 2 * 2 = 5 Example 2: Input: 3 Output: False 这道题让我们求一个数是否能由平方数之和组成,刚开始博主没仔细看题,没有看…

题目: Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c. Example 1: Input: 5 Output: True Explanation: 1 * 1 + 2 * 2 = 5 Example 2: Input: 3 Output: False 分析: 给定一个非负整数c ,你要判断是否存在两个整数a和b,使…

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…

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…

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…

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   /  …

题目 404. 左叶子之和 如题 题解 类似树的遍历的递归 注意一定要是叶子结点 代码 class Solution { public int sumOfLeftLeaves(TreeNode root) { if(root == null){return 0;} int sum = sumOfLeftLeaves(root.left)+sumOfLeftLeaves(root.right); if(root.left!=null&&root.left.left==null&&am…

404. 左叶子之和 知识点:二叉树 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 解法一:DFS 函数功能:左叶子之和 1.终止条件:root为空,返回0: 2.能做什么:判断自己的左节点是否为空,不为空的话判断它是不是叶子节点,是的话就加到sum上:不是的话那就接着去看子树: 3.什么时候做:从上到下,先弄自己的,再去弄子树的,前序: 做这类二叉树的题目,多半是遍历树,遍历的过程…

404. 左叶子之和 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 /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solutio…

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 解析 我们需要找到这样的节点 属于叶子节点 属于父节点的左子节点 方法一:用栈,dfs遍历,用全局变量res作为累积和.遍历的过程中传递该节点是否是左子节点.同时判断左右子节点是否为None,则可以知道是不是左叶子节点. class Solution: def sumOfLeftLeaves(self, root: TreeNode) -> int…

LeetCode:Path Sum Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \…

Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target. Example 1: Input: 5 / \ 3 6 / \ \ 2 4 7 Target = 9 Output: True Example 2: Input: 5 / \ 3 6 / \ \ 2 4…

剑指offer 65. 不用加减乘除做加法(Leetcode 371. Sum of Two Integers) https://leetcode.com/problems/sum-of-two-integers/ 题目: 写一个函数,求两个整数之和,要求在函数体内不得使用加减乘除这四个符号. 分析: 对于不能使用正常的四则运算符,一般就是使用位运算了.而本题要想实现加法,只能使用异或了. 需要注意的是,加法的时候涉及进位,而进位的实现利用与运算. 此外,进位之后还有可能产生进位,所以要在循环里…