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解法一：（深度优先搜索）穷举所有的可能 =》访问每一个节点 node，检测以 node 为起始节点且向下延深的路径有多少种。递归遍历每一个节点的所有可能的路径，然后将这些路径数目加起来即为返回结果。

/*** 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, int targetSum) {if(root==null){return 0;}int res = rootSum(root, targetSum);res += pathSum(root.left, targetSum);res += pathSum(root.right, targetSum);return res;}public int rootSum(TreeNode root, int targetSum){if(root==null){return 0;}int res = 0;if(root.val==targetSum){res++;}res += rootSum(root.left, targetSum-root.val);res += rootSum(root.right, targetSum-root.val);return res;}
}

注意：

• public int pathSum(TreeNode root, int targetSum)：根节点+左节点+右节点分开单独遍历每一个节点的所有可能的路径；public int rootSum(TreeNode root, int targetSum)遍历根节点所有可能的路径。
• 将pathSum(TreeNode root, int targetSum)和rootSum(TreeNode root, int targetSum)的int改为long可以解决下述问题。