Path Sum III

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.

题意：找出所以路径和为sum的路径总个数，不要求起点为根节点也不要求终点为叶子节点，只有一个要求：就是必须是从上往下遍历

解析：这题目标着easy，思想也确实简单，能想到用dfs遍历即可，那它既然不要求起点为根节点，就是说需要对每个节点都进行一次dfs。但是虽然知道这个思路，但是我却没能很好的写出来，看了solution才恍悟可以这么去写。

//pathSum表示对一个节点的DFS//solver表示dfs的递归式    public int pathSum(TreeNode root, int sum) {        if (root == null) return 0;        return soler(root,sum) + pathSum(root.left,sum) + pathSum(root.right,sum);    }    private int soler(TreeNode root,int sum) {        if (root == null)            return 0;        return (root.val == sum? 1 : 0) + soler(root.left,sum - root.val) + soler(root.right,sum - root.val);    }