res += tmp[sum]; res +=pathSum(root->left, sum, tmp); res +=pathSum(root->right, sum, tmp); }returnres; }intpathSum(TreeNode* root,intsum){ unordered_map<longlong,longlong> mp; mp[sum] =0;returnpathSum(root, sum, mp); } };...
https://leetcode.com/problems/path-sum-iii/discuss/141424/Python-step-by-step-walk-through.-Easy-to-understand.-Two-solutions-comparison.-:-) 特别需要注意,hashtable保存的路径和,一定是能延伸到当前节点的。所以函数最后 --count[curSum] 是必须加的,当递归返回上一层时,包含当前节点的路径 curSum ...
递归左右节点时,这时候的sum值应该是原sum值减去当前节点的值。 class Solution: def hasPathSum(self, root, sum): """ :type root: TreeNode :type sum: int :rtype: bool """ if root is None: return False if (root.val == sum) and (root.left is None) and (root.right is None): re...
public class Solution private int _sum; private int _count; public int PathSum(TreeNode root, int sum) _count = 0; _sum = sum; Travel(root, new List<int>()); return _count; private void Travel(TreeNode current, List<int> ret) if(current == null) return ; if(current.val == _...
LeetCode Top 100 Liked Questions 437. Path Sum III (Java版; Easy) 题目描述 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 ...
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
Left, remainSum) { return true } // 当右子结点存在,且存在一条右子结点到叶子路径上所有值到和为 remainSum ,则满足题意 if root.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...
returnresult+[path+[root.val]] else: returnself._pathSum(root.left,sum-root.val,path+[root.val],result)+self._pathSum(root.right,sum-root.val,path+[root.val],result) Leetcode 笔记系列的Python代码共享在https://github.com/wizcabbit/leetcode.solution...
931. 下降路径最小和 - 给你一个 n x n 的 方形 整数数组 matrix ,请你找出并返回通过 matrix 的下降路径 的 最小和 。 下降路径 可以从第一行中的任何元素开始,并从每一行中选择一个元素。在下一行选择的元素和当前行所选元素最多相隔一列(即位于正下方或者沿对角线向
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) {} * }; */class Solution{public:intpathSum(TreeNode*root,intsum){if(!roo...