标签：lock from etc style path sum post hat == .com
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
class Solution { public: int pathSum(TreeNode* root, int sum) { if (root == nullptr) return 0; return dfs(root, sum) + pathSum(root->left, sum) + pathSum(root->right, sum); } int dfs(TreeNode* node, int sum) { if (node == nullptr) return 0; return (node->val == sum ? 1 : 0) + dfs(node->left, sum - node->val) + dfs(node->right, sum - node->val); } }; // 22 ms
标签：lock from etc style path sum post hat == .com
原文地址：http://www.cnblogs.com/immjc/p/7231524.html
