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>> result;
        if(!root){
            return result;
        }
        
        vector<int> answer;
        backtrace(root, sum, answer, result);
        return result;
    }
    
    void backtrace(TreeNode* root, int sum, vector<int> answer, vector<vector<int>>& result){
        //assert(root);
        if(root && root->val == sum && !root->left && !root->right){
            answer.push_back(root->val);
            result.push_back(answer);
        }
        
        answer.push_back(root->val);
        if(root->left){
            backtrace(root->left, sum-root->val, answer, result);
        }
        
        if(root->right){
            backtrace(root->right, sum-root->val, answer, result);
            
        }
        answer.pop_back();
    }
};