java 输出二叉树的所有跟节点到叶子节点的路径

给定一个二叉树，返回所有从根节点到叶子节点的路径。

说明: 叶子节点是指没有子节点的节点。

示例: 输入: 

1

/ \

2 3

\

5

输出: ["1->2->5", "1->3"]  

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

package ds;

import java.util.LinkedList;
import java.util.List;

public class AllNodePath {

    private static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        public TreeNode(int val) {
            this.val = val;
        }
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(6);
        TreeNode node11 = new TreeNode(2);
        TreeNode node12 = new TreeNode(7);
        TreeNode node21 = new TreeNode(1);
        TreeNode node22 = new TreeNode(5);
        TreeNode node31 = new TreeNode(3);
        TreeNode node42 = new TreeNode(4);
        root.left = node11;
        root.right = node12;
        node11.left = node21;
        node11.right = node22;
        node22.left = node31;
        node31.right = node42;
        List res=binaryTreePaths(root);
        System.out.println(res);
    }

    public static List binaryTreePaths(TreeNode root) {
        LinkedList res = new LinkedList<>();
        if(root == null) return res;
        solve(root, "", res);
        return res;
    }

    public static void solve(TreeNode root, String cur, LinkedList res) {
        if (root == null) {
            return;
        }
        cur += root.val;
        if (root.left == null && root.right == null) {
            res.add(cur);
        } else {
            solve(root.left, cur + "->", res);
            solve(root.right, cur + "->", res);
        }

    }
}

第二种解决方法

package ds;

import java.util.LinkedList;

public class AllNodePath {
    static LinkedList> res=new LinkedList<>();

    private static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        public TreeNode(int val) {
            this.val = val;
        }
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(6);
        TreeNode node11 = new TreeNode(2);
        TreeNode node12 = new TreeNode(7);
        TreeNode node21 = new TreeNode(1);
        TreeNode node22 = new TreeNode(5);
        TreeNode node31 = new TreeNode(3);
        TreeNode node42 = new TreeNode(4);
        root.left = node11;
        root.right = node12;
        node11.left = node21;
        node11.right = node22;
        node22.left = node31;
        node31.right = node42;
        LinkedList ll=new LinkedList<>();
        getAllPaths(root,ll);
        System.out.println(res);
    }

    private static void getAllPaths(TreeNode root,LinkedList ll){
        if(root==null){
            return ;
        }
        ll.add(root.val);
        if(root.left==null&&root.right==null){
            LinkedList temp=new LinkedList<>();
            for(int i=0;i