PRoblem:Given a binary tree, return all root-to-leaf paths.

Solution:每個(gè)結(jié)點(diǎn)記錄從root至其的path，當(dāng)結(jié)點(diǎn)無(wú)左右孩子時(shí)，將path傳入進(jìn)vector。

這里需要注意的是，recordPath函數(shù)中，path記錄著root至結(jié)點(diǎn)的path，不可改變，不能加&（reference），而pathResult需要添加新的path，所以要加&。

void recordPath(TreeNode* tn,vector<int> path,vector< vector<int> >& pathResult)
class Solution {    void recordPath(TreeNode* tn,vector<int> path,vector< vector<int> >& pathResult){        path.push_back(tn->val);        if(!tn->left&&!tn->right)            pathResult.push_back(path);        if(tn->left)            recordPath(tn->left,path,pathResult);        if(tn->right)            recordPath(tn->right,path,pathResult);        return;    }    void convert(ve,vector< vector<int> > path)ctor<string>& result{
        int rowSize=path.size();        for(int i=0;i<rowSize;i++){            string tmp;            tmp=to_string(path[i][0]);            for(int j=1;j<path[i].size();j++){                tmp+="->";                tmp+=to_string(path[i][j]);            }            result.push_back(tmp);        }        return;    }public:    /**     * @param root the root of the binary tree     * @return all root-to-leaf paths     */    vector<string> binaryTreePaths(TreeNode* root) {        vector<string> result;        vector< vector<int> > pathResult;        vector<int> tmp;        if(!root)            return result;        recordPath(root,tmp,pathResult);        convert(result,pathResult);        return result;    }};