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   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

这道题让我们求二叉树的路径的和等于一个给定值，说明了这条路径不必要从根节点开始，可以是中间的任意一段，而且二叉树的节点值也是有正有负。那么我们可以用递归来做，相当于先序遍历二叉树，对于每一个节点都有记录了一条从根节点到当前节点到路径，同时用一个变量curSum记录路径节点总和，然后我们看curSum和sum是否相等，相等的话结果res加1，不等的话我们来继续查看子路径和有没有满足题意的，做法就是每次去掉一个节点，看路径和是否等于给定值，注意最后必须留一个节点，不能全去掉了，因为如果全去掉了，路径之和为0，而如果假如给定值刚好为0的话就会有问题，整体来说不算一道很难的题，参见代码如下：

解法一：

class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        int res = 0;
        vector<TreeNode*> out;
        helper(root, sum, 0, out, res);
        return res;
    }
    void helper(TreeNode* node, int sum, int curSum, vector<TreeNode*>& out, int& res) {
        if (!node) return;
        curSum += node->val;
        out.push_back(node);
        if (curSum == sum) ++res;
        int t = curSum;
        for (int i = 0; i < out.size() - 1; ++i) {
            t -= out[i]->val;
            if (t == sum) ++res;
        }
        helper(node->left, sum, curSum, out, res);
        helper(node->right, sum, curSum, out, res);
        out.pop_back();
    }
};

我们还可以对上面的方法进行一些优化，来去掉一些不必要的计算，我们可以用哈希表来建立所有的前缀路径之和跟其个数之间的映射，然后看子路径之和有没有等于给定值的，参见代码如下：

解法二：

class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        unordered_map<int, int> m;
        m[0] = 1;
        return helper(root, sum, 0, m);
    }
    int helper(TreeNode* node, int sum, int curSum, unordered_map<int, int>& m) {
        if (!node) return 0;
        curSum += node->val;
        int res = m[curSum - sum];
        ++m[curSum];
        res += helper(node->left, sum, curSum, m) + helper(node->right, sum, curSum, m);
        --m[curSum];
        return res;
    }
};

下面这种方法非常的简洁，也是利用了前序遍历，对于每个遍历到的节点进行处理，维护一个变量pre来记录之前路径之和，然后cur为pre加上当前节点值，如果cur等于sum，那么返回结果时要加1，然后对当前节点的左右子节点调用递归函数求解，参见代码如下：

解法三：

class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        if (!root) return 0;
        return sumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
    }
    int sumUp(TreeNode* node, int pre, int& sum) {
        if (!node) return 0;
        int cur = pre + node->val;
        return (cur == sum) + sumUp(node->left, cur, sum) + sumUp(node->right, cur, sum);
    }
};

类似题目：

Binary Tree Maximum Path Sum

Path Sum II

Path Sum

Minimum Path Sum

参考资料：

https://discuss.leetcode.com/topic/64388/simple-ac-java-solution-dfs/2

https://discuss.leetcode.com/topic/64402/c-5-line-body-code-dfs-solution