题目描述：

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

For example, given the following binary tree:

1 /   \2     3 \  5

All root-to-leaf paths are:

["1->2->5", "1->3"]

看到树相关的题目，第一反应就是递归，下面是我的代码：

vector
     
       binaryTreePaths(TreeNode* root) {    vector
      
        res;    if (root != NULL)    {        char buf[20];        sprintf(buf, "%d", root->val);        string rootPath = buf;        if (root->left == NULL && root->right == NULL)        {            res.push_back(rootPath);        }        else        {            if (root->left)            {                vector
       
         leftPath;                leftPath = binaryTreePaths(root->left);                for (vector
        
         ::iterator it = leftPath.begin(); it != leftPath.end(); ++it)                {                    res.push_back(rootPath + "->" + *it);                }            }            if (root->right)            {                vector
         
           rightPath;                rightPath = binaryTreePaths(root->right);                for (vector
          
           ::iterator it = rightPath.begin(); it != rightPath.end(); ++it) { res.push_back(rootPath + "->" + *it); } } } } return res;}
          
         
        
       
      
     

想要简洁一点的，可以参考