437. Path Sum III（路径可以任意点开始，任意点结束）

【437. Path Sum III（路径可以任意点开始，任意点结束）】的更多相关文章

[LeetCode] 437. Path Sum III 路径和 III

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…

leetcode 437 Path Sum III 路径和

相关问题:112 path sum /** * 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 pathSum(TreeNode* root, int s…

437 Path Sum III 路径总和 III

给定一个二叉树,二叉树的每个节点含有一个整数.找出路径和等于给定数的路径总数.路径不需要从根节点开始,也不需要在叶节点结束,当路径方向必须是向下的(只从父节点到子节点).二叉树不超过1000个节点,节点的整数值的范围是[-1000000,1000000].示例:root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 10 / \ 5 -3 / \ \ 3 2 11 / \ \3 -2 1返回…

47. leetcode 437. Path Sum III

437. Path Sum III 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 fro…

437. Path Sum III

原题: 437. Path Sum III 解题: 思路1就是:以根节点开始遍历找到适合路径,以根节点的左孩子节点开始遍历,然后以根节点的右孩子节点开始遍历,不断循环,也就是以每个节点为起始遍历点代码如下: class Solution { public: int pathSum(TreeNode* root, int sum) { if(!root) return 0; int count = getPath(root,0,sum) + pathSum(root->left,sum) + p…

【leetcode】437. Path Sum III

problem 437. Path Sum III 参考 1. Leetcode_437. Path Sum III; 完…

leetcode 112. Path Sum 、 113. Path Sum II 、437. Path Sum III

112. Path Sum 自己的一个错误写法: class Solution { public: bool hasPathSum(TreeNode* root, int sum) { if(root == NULL) return false; ; return hasPathSum(root,sum,value); } bool hasPathSum(TreeNode* root,int sum,int value){ if(root == NULL){ if(value == sum) r…