437. 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.

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 -> 1
3. -3 -> 11
思路：用分治的思想做，得到left的pathsum,right的pathsum，再加上当前节点的sum值，进行返回。当计算当前的sum值时，还需要一个递归：
/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */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);    }    private int dfs(TreeNode root, int sum) {        if(root == null){            return 0;        }        if(root.val == sum){            return 1 + dfs(root.left, sum - root.val) + dfs(root.right, sum - root.val);        }        return dfs(root.left, sum - root.val) + dfs(root.right, sum - root.val);    }}