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 no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ 5 -3
/ \ 3 2 11
/ \ 3 -2 1
Return 3. The paths that sum to 8 are:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
思路：深度优先搜索，分别以root, root->left, root->right作为开始节点，计算下方是否有满足和为sum的路径。dfs用于统计个数。
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 void dfs(TreeNode* root, int sum, int &num) { 12 if (root == NULL) 13 return; 14 if (sum == root->val) 15 num++; 16 dfs(root->left, sum - root->val, num); 17 dfs(root->right, sum - root->val, num); 18 } 19 public: 20 int pathSum(TreeNode* root, int sum) { 21 if (root == NULL) { 22 return 0; 23 } 24 int num = 0; 25 dfs(root, sum, num); 26 int left = pathSum(root->left, sum); 27 int right = pathSum(root->right, sum); 28 return num + left + right; 29 } 30 };
原文：https://www.cnblogs.com/qinduanyinghua/p/11431460.html