给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。
说明: 叶子节点是指没有子节点的节点。
示例:
给定如下二叉树，以及目标和 sum = 22，5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
返回:[
[5,4,11,2],
[5,8,4,5]
]
解题思路：
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:vector<vector<int>> pathSum(TreeNode* root, int sum) {if(!root) return {};vector<vector<int>> res;dfs(root, sum, {}, res);return res;}void dfs(TreeNode *root, int sum, vector<int> curr, vector<vector<int>> &res){if(!root) return;curr.push_back(root->val);if(root->left == NULL && root->right == NULL){if(root->val == sum)res.push_back(curr);return;}dfs(root->left, sum - root->val, curr, res);dfs(root->right, sum - root->val, curr, res);}
};
)
)
