res +=pathSum(root->right, sum, tmp); }returnres; }intpathSum(TreeNode* root,intsum){ unordered_map<longlong,longlong> mp; mp[sum] =0;returnpathSum(root, sum, mp); } };
returndfs(root,sum)+self.pathSum(root.left,sum)+self.pathSum(root.right,sum) 尤其注意是dfs(root, sum)+self.pathSum(root.left, sum)+self.pathSum(root.right, sum) 而不是dfs+dfs+dfs!容易出错! 另外解法就是前缀树的解法,前缀树记录sum,比如: root = [10,5,-3,3,2,null,11,3,-2,null...
You are given a binary tree in which each node contains an integer value. Find the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes). The tree has ...
AI检测代码解析 packageleetcode.easy;/** * Definition for a binary tree node. public class TreeNode{int val; TreeNode * left; TreeNode right; TreeNode(int x){val = x;}}*/publicclassPathSumIII{intcount=0;privatevoidhelper(TreeNoderoot,intsum){if(root==null){return;}if(root.val==sum...
Right != nil && hasPathSum(root.Right, remainSum) { return true } // 左右子结点都不满足题意,返回 false return false } 题目链接: Path Sum: leetcode.com/problems/p 路径总和: leetcode.cn/problems/pa LeetCode 日更第 262 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
leetcode 112. Path Sum 、 113. Path Sum II 、437. Path Sum III,112.PathSum自己的一个错误写法:只有左右节点都为NULL时才是叶子节点,所以这个代码在例子[1,2],1的右节点时就判断错误了,这个右节点虽然sum满足条件,但他本身不是叶子节点正确写法:113.PathSumII第
【CSON】LeetCode讲解 64. Minimum Path Sum发布于 2022-01-13 11:41 · 274 次播放 赞同添加评论 分享收藏喜欢 举报 力扣(LeetCode)算法编程程序员面试移动社交 写下你的评论... 还没有评论,发表第一个评论吧相关推荐 10:51 大型MMORPG「我们结婚吧」,玩家流失严重 #零基础看懂...
题目链接:Path Sum II | LeetCode OJ Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22, 5 /\ 48 //\ 11134 /\/\ ...
1. Description Path Sum III 2. Solution Recursive /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
path(root.left,sum); path(root.right,sum); } } 113. 路径总和 II 给定一个二叉树和一个目标和,找到所有从根节点到叶子节点路径总和等于给定目标和的路径。 说明:叶子节点是指没有子节点的节点。 示例: 给定如下二叉树,以及目标和sum = 22,