[113] 路径总和 II

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:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> res;
        if(root == NULL){
            return res;
        }
        sum = sum - root->val;
        if(root->left == NULL && root->right == NULL){
            if(sum == 0){
                vector<int> temp;
                temp.push_back(root->val);
                res.push_back(temp);
                return res;
            }else {
                return res;
            }
        }
        vector<vector<int>> left  = pathSum(root->left, sum);
        vector<vector<int>> right = pathSum(root->right, sum);
        if(left.size() > 0){
            for(int i = 0; i<left.size();i++){
                vector<int> temp = left[i];
                temp.insert(temp.begin(), root->val);
                res.push_back(temp);
            }
        }

        if(right.size() > 0){
            for(int i = 0; i<right.size();i++){
                vector<int> temp = right[i];
                temp.insert(temp.begin(), root->val);
                res.push_back(temp);
            }
        }

        return  res;
    }
};