LeetCode OJ：Binary Tree Paths（二叉树路径）

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"]

简单的遍历查找路径问题，代码如下：

 /**
  * 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) {
         ret.clear();
         string s = "";
         if(root == NULL) return ret;
         dfs(root, s);
         for(int i = ; i < ret.size(); ++i){
             ret[i].erase(ret[i].begin(), ret[i].begin() + );
         }
         return ret;
     }

     void dfs(TreeNode * root, string s)
     {
         stringstream ss;
         ss << "->" << root->val;
         s += ss.str();
         if(root->left == NULL && root->right == NULL){
             ret.push_back(s);
             return;
         }
         if(root->left){
             dfs(root->left, s);
         }
         if(root->right){
             dfs(root->right, s);
         }
     }
 private:
     vector<string> ret;
 };

java版本的如下所示，和c++的相比还是要简单很多的，因为处理字符串的函数用起来比较方便的原因，代码如下：

 public class Solution {
     public List<String> binaryTreePaths(TreeNode root) {
         List<String> ret = new ArrayList<String>();
         String str = new String();
         if(root == null)
             return ret;
         dfs(root, str, ret);
         return ret;
     }
     public void dfs(TreeNode root, String path, List<String> ret){
         if(root.left != null){
             path = path + "->" + root.val;
             dfs(root.left, path, ret);
             path = path.substring(0, path.lastIndexOf("->")); //引用其他的
         }                                        //还是要继续使用的，截断即可
         if(root.right != null){
             path = path + "->" + root.val;
             dfs(root.right, path, ret);
             path = path.substring(0, path.lastIndexOf("->"));
         }
         if(root.left == null && root.right == null){
             path = path + "->" + root.val;
             ret.add(path.substring(2));
             path = path.substring(0,path.lastIndexOf("->"));
             return;
         }
     }
 }