[leetcode] 437. Path Sum III

Question:

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   1Return 3. The paths that sum to 8 are:1.  5 -> 32.  5 -> 2 -> 13. -3 -> 11

Solution:

直接DFS，算是一种暴力求解方法，在pathSum中尝试以每个节点作为根节点往下遍历。在help中不断往下遍历试图找到val==sum的节点，注意在help中sum会随着往下遍历一直变化。
从遍历的次数来看，肯定存在重复冗余的遍历，就是不知道要怎么做才能以空间换时间。

/** * 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 help(root, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);    }    int help(TreeNode* root, int sum) {        if (root == NULL) return 0;        int count = 0;        return (root->val == sum ? 1 : 0) + help(root->left, sum - root->val) + help(root->right, sum - root->val);    }};