树、递归————二叉树的所有路径

递归终止条件：到达叶子节点，即该节点的左右孩子均为空。 
路径的拼接：每个节点值 + -> + 左孩子的路径/右孩子的路径（左右均可能有多个路径）

/**
 * 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<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if(root == NULL) return res;
        if(root->left == NULL && root->right == NULL)
            res.push_back(to_string(root->val));
        //不是叶子节点的话
        vector<string> leftS = binaryTreePaths(root->left);
        for(int i = 0; i < leftS.size(); i ++)
            res.push_back(to_string(root->val) + "->" + leftS[i]);
        vector<string> rihgtS = binaryTreePaths(root->right);
        for(int i = 0; i < rihgtS.size(); i ++)
            res.push_back(to_string(root->val) + "->" + rihgtS[i]);
        return res;
    }
};