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LeetCode : Path Sum III

2017-03-15 16:14 435 查看
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;
}
};
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标签:  leetcode