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]
]

/**
* 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:
void sumVec(vector<vector<int>>&res, TreeNode* root, vector<int> temp, int expectNumber, int add)
{
if(root==NULL)
return;
temp.push_back(root->val);
int x=root->val+add;
if(root->left==NULL && root->right==NULL && x==expectNumber)
res.push_back(temp);
sumVec(res, root->left, temp, expectNumber, x);
sumVec(res, root->right, temp, expectNumber, x);
}
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> res;
vector<int> temp;
sumVec(res, root, temp, sum, 0);
return res;
}
};