LeetCode : Path Sum III

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
/**
 * 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;
    }
};
原文地址:https://www.cnblogs.com/chankeh/p/6850092.html