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

public class liubobo_7_5 {
    /// 437. Path Sum III
/// https://leetcode.com/problems/path-sum-iii/description/
/// 时间复杂度: O(n), n为树的节点个数
/// 空间复杂度: O(h), h为树的高度
        /// Definition for a binary tree node.
        public static class TreeNode {
            int val;
            TreeNode left;
            TreeNode right;
            TreeNode(int x) { val = x; }
        }

        // 在以root为根节点的二叉树中,寻找和为sum的路径,返回这样的路径个数
        public int pathSum(TreeNode root, int sum) {

            if(root == null)
                return 0;

            return findPath(root, sum)
                    + pathSum(root.left , sum)
                    + pathSum(root.right , sum);
        }

        // 在以node为根节点的二叉树中,寻找包含node的路径,和为sum
        // 返回这样的路径个数
        private int findPath(TreeNode node, int num){

            if(node == null)
                return 0;

            int res = 0;
            if(node.val == num)
                res += 1;

            res += findPath(node.left , num - node.val);
            res += findPath(node.right , num - node.val);

            return res;
        }

        public static void main(String[] args) {

            // 手动创建Leetcode题页上的测试用例。
            // 当然, 有更好的更智能的创建二叉树的方式, 有兴趣的同学可以自行研究编写程序:)

            /*****************
             * 测试用例:
             *
             *       10
             *      /  \
             *     5   -3
             *    / \    \
             *   3   2   11
             *  / \   \
             * 3  -2   1
             *****************/
            TreeNode node1 = new TreeNode(3);
            TreeNode node2 = new TreeNode(-2);

            TreeNode node3 = new TreeNode(3);
            node3.left = node1;
            node3.right = node2;

            TreeNode node4 = new TreeNode(1);
            TreeNode node5 = new TreeNode(2);
            node5.right = node4;

            TreeNode node6 = new TreeNode(5);
            node6.left = node3;
            node6.right = node5;

            TreeNode node7 = new TreeNode(11);
            TreeNode node8 = new TreeNode(-3);
            node8.right = node7;

            TreeNode node9 = new TreeNode(10);
            node9.left = node6;
            node9.right = node8;

            System.out.println((new liubobo_7_5()).pathSum(node9, 8));
        }
    }