[Leetcode] 437. Path Sum III 解题报告

题目：

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

思路：

虽然在Leetcode中这道题目的level是easy，但我感觉要写出优雅的代码并不容易（至少我刚开始写的代码就一塌糊涂）。这里关键在于我们要覆盖掉所有情况，并且不重复计算。思路是：对于以root为根的树来说，我们可以将符合条件的path分为两类，一类是包含root的，一类是不包含root的。对于包含root的，我们用DFS函数来求解；而对于不包含root的path来说，我们就可以把它分为左右两棵子树来分别进行求解。最终的结果就是三者之和。下面的代码片段给出了具体实现，并且附加了详细解释。

代码：

/** * 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 ret = pathSum(root->left, sum) + pathSum(root->right, sum);     // root not included in the path        return DFS(root, sum) + ret;                                        // root included in the path    }private:    int DFS(TreeNode* root, int sum) {                  // the numer of paths that begin from root        if (!root) {            return 0;        }        int ret = 0;        if (sum == root->val) {                         // the path that ends here            ++ret;        }        ret += DFS(root->left, sum - root->val);        // the paths that end somewhere in the left sub tree        ret += DFS(root->right, sum - root->val);       // the paths that end somewhere in the right sub tree        return ret;    }};