Level：
Easy
题目描述：
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 = 810/ \5 -3/ \ \3 2 11/ \ \
3 -2 1Return 3. The paths that sum to 8 are:1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11思路分析：
由于题目中所说的路径，不局限于从根到叶子节点，任何一个节点都可以作为路径的开始节点和终止节点，所以我们以根节点作为开始节点，查找和为sum的路径数，然后分别以根的左孩子和右孩子为起始节点去查找和为sum的路径数，依次递归向下推导，得到最终的结果。
代码：
/**public class TreeNode{int vla;TreeNode left;TreeNode right;public TreeNode(int val){this.val=val;}
}*/
public class Soulution{public int pathSum(TreeNode root,int sum){if(root==null)return 0;int res=0;res=pathCheck(root,sum);res=res+pathSum(root.left,sum);res=res+pathSum(root.right,sum);return res;}public int pathCheck(TreeNode root,int sum){if(root==null)return 0;int count=0;if(sum==root.val)//当sum等于root.val时证明存在一条路径和为sumcount++;count=count+pathCheck(root.left,sum-root.val);count=count+pathCheck(root.right,sum-root.val);return count;}
} 
