leetcode437.路径总和|||

对于根结点来说，可以选择当前结点为路径也可以不选择，但是一旦选择当前结点为路径那么后续都必须要选择结点作为路径，不然路径不连续是不合法的，所以这里分开出来两个方法进行递归

由于力扣最后一个用例解答错误，分析发现targetSum减法多次后可能越界之类的情况把参数类型改为了long

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int pathSum(TreeNode root, long targetSum) {
        if (root == null) {
            return 0;
        } else {
            if (targetSum - root.val == 0) {
                return 1
                        + getSum(root.left, targetSum - root.val)
                        + getSum(root.right, targetSum - root.val)
                        + pathSum(root.left, targetSum)
                        + pathSum(root.right, targetSum);
            } else {
                return getSum(root.left, targetSum - root.val)
                        + getSum(root.right, targetSum - root.val)
                        + pathSum(root.left, targetSum)
                        + pathSum(root.right, targetSum);
            }
        }
    }

    private int getSum(TreeNode root, long targetSum) {
        if (root == null) {
            return 0;
        } else {
            targetSum-=root.val;
            if (targetSum == 0) {
                return 1 + getSum(root.left, targetSum) + getSum(root.right, targetSum);
            } else {
                return getSum(root.left, targetSum) + getSum(root.right, targetSum);
            }
        }
    }
}