113. Path Sum II 这里要注意 引用 和 传值的区别，见注释

题目：

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

For 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]]

代码和思路：

vector<vector<int>> pathSum(TreeNode* root, int sum) {        vector<int> fa;        vector<vector<int>> ans;        help(root, sum, fa, ans );        return ans;    }

下面是传值：

void help(TreeNode* root, int sum, vector<int> &father, vector<vector<int>>& ans)    {        if(root == NULL) return ;        vector<int> temp(father);        temp.push_back(root->val);        if(root->left == NULL && root->right == NULL){            if(root->val == sum){                ans.push_back(temp);            }            return ;        }        help(root->left, sum - root->val, temp, ans);        help(root->right, sum - root->val, temp, ans);    }

相当于：

void help(TreeNode* root, int sum, vector<int> father, vector<vector<int>>& ans)    {        if(root == NULL) return ;        /*vector<int> temp(father);        temp.push_back(root->val);*/                father.push_back(root->val);        if(root->left == NULL && root->right == NULL){            if(root->val == sum){                ans.push_back(father);            }            return ;        }        help(root->left, sum - root->val, father, ans);        help(root->right, sum - root->val, father, ans);    }

而对于引用传递的话一定要进行pop操作，注意体会：这里就是回溯法

void help(TreeNode* root, int sum, vector<int> &father, vector<vector<int>>& ans)    {        if(root == NULL) return ;                father.push_back(root->val);        if(root->left == NULL && root->right == NULL){            if(root->val == sum){                ans.push_back(father);            }            //return ;这里return一定要去掉 注意体会 因为在下面两个help中若找不到（没有root->val == sum）那么就要pop_back.就是一种回溯法        }        help(root->left, sum - root->val, father, ans);        help(root->right, sum - root->val, father, ans);        father.pop_back();    }

最后一个help只要9msAC，说明引用传递效率要高。