對于根結點來說，可以選擇當前結點為路徑也可以不選擇，但是一旦選擇當前結點為路徑那么后續都必須要選擇結點作為路徑，不然路徑不連續是不合法的，所以這里分開出來兩個方法進行遞歸

由于力扣最后一個用例解答錯誤，分析發現targetSum減法多次后可能越界之類的情況把參數類型改為了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);}}}
}

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