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
/**
* 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;
}
};