leetcode 257. Binary Tree Paths 深度优先遍历DFS

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”]

直接DFS深度优先遍历即可。

代码如下：

import java.util.ArrayList;import java.util.List;/*class TreeNode{     int val;     TreeNode left;     TreeNode right;     TreeNode(int x) { val = x; }}*//* * 这道题其实很简单，但是要注意递归开始的起点和终点 * Java中使用StringBuilder会更快点 * */public class Solution {    List<String> res=new ArrayList<String>();    public List<String> binaryTreePaths(TreeNode root)    {        if(root==null)            return res;        //getAllPath(root,root.val+"");        List<Integer> one=new ArrayList<>();        one.add(root.val);        getAllPath(root, one);        return res;    }    private void getAllPath(TreeNode root, List<Integer> one)    {        if(root.left==null && root.right==null)        {            if(one.size()==1)                res.add(one.get(0)+"");            else             {                StringBuilder lastBuilder=new StringBuilder();                lastBuilder.append(one.get(0));                for(int i=1;i<one.size();i++)                    lastBuilder.append("->"+one.get(i));                res.add(lastBuilder.toString());            }        }        if(root.left!=null)        {               one.add(root.left.val);            getAllPath(root.left, one);            one.remove(one.size()-1);        }           if(root.right!=null)        {               one.add(root.right.val);            getAllPath(root.right, one);            one.remove(one.size()-1);        }    }    private void getAllPath(TreeNode root, String path)    {        if(root.left==null && root.right==null)            res.add(path);        if(root.left!=null)            getAllPath(root.left, path+"->"+root.left.val);        if(root.right!=null)            getAllPath(root.right, path+"->"+root.right.val);    }}

下面是C++的做法，就是做一个简单的DFS深度优先遍历，代码如下：

#include <iostream>#include <vector>#include <string>using namespace std;/*struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {}};*/class Solution {public:    vector<string> res;    vector<string> binaryTreePaths(TreeNode* root)     {        if (root == NULL)            return res;        getAll(root, to_string(root->val));        return res;    }    void getAll(TreeNode* root, string one)    {        if (root->left == NULL && root->right == NULL)            res.push_back(one);        if (root->left != NULL)            getAll(root->left, one + "->" + to_string(root->left->val));        if (root->right != NULL)            getAll(root->right, one + "->" + to_string(root->right->val));    }};