算法描述：

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

解题思路：回溯法。注意确认边界条件。

    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> results;
        vector<int> temp;
        backtracking(root, results, temp, sum);
        return results;
    }
    
    void backtracking(TreeNode* root, vector<vector<int>>& results, vector<int>& temp, int sum){
        if(root==nullptr) return;
        temp.push_back(root->val);
        if(sum ==root->val && root->left==nullptr && root->right==nullptr){
            results.push_back(temp);
        } 
        backtracking(root->left, results, temp, sum-root->val);
        backtracking(root->right, results, temp, sum-root->val);
        temp.pop_back();
    }