【二叉树】437. Path Sum III(理解递归)

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

要注意，题目说了路径可以不从根开始

解答：

/**
* 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) return 0;
int cur=0;
return dfs(root,sum,0)+pathSum(root->left,sum)+pathSum(root->right,sum);
}
int dfs(TreeNode* root,int sum,int pre){
if(!root) return 0;
int cur=pre+root->val;
return (cur==sum)+dfs(root->left,sum,cur)+dfs(root->right,sum,cur);
}
};这个题的递归还是很有意思的，递归的题目就是。。写出来觉得很简单，写的时候却总出错。。。