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 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:

5 -> 3

5 -> 2 -> 1

-3 -> 11

这道题就是递归的方法，先从根部开始去递归计算每一个子树是否存在和为sum的路径，sum-root->root如果结果等于下一个节点的val，那么这条路径符合条件。计数器为res。然后是在pathSum函数中还要通过把左右子节点作为根节点再去找可能的路径，把结果相加即可。

代码如下：

/**
* 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 sum) {
if(root==NULL){
return 0;
}
return dfs(root,sum)+pathSum(root->left,sum)+pathSum(root->right,sum);
}
int dfs(TreeNode* root,int sum){
int res=0;
if(root==NULL){
return res;
}
if(sum==root->val) ++res;
res+=dfs(root->left,sum-root->val);
res+=dfs(root->right,sum-root->val);
return res;
}
};