9.path Sum III（路径和 III）

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 = 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

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sum
            count++;
        count=count+pathCheck(root.left,sum-root.val);
        count=count+pathCheck(root.right,sum-root.val);
        return count;
    }
}