//给你一个二叉树的根节点 root ，按 任意顺序 ，返回所有从根节点到叶子节点的路径。 
//
// 叶子节点 是指没有子节点的节点。 
//
// 示例 1： 
// 
// 
//输入：root = [1,2,3,null,5]
//输出：["1->2->5","1->3"]
// 
//
// 示例 2： 
//
// 
//输入：root = [1]
//输出：["1"]
// 
//
// 
//
// 提示： 
//
// 
// 树中节点的数目在范围 [1, 100] 内 
// -100 <= Node.val <= 100 
// 
//
// Related Topics 树 深度优先搜索 字符串 回溯 二叉树 👍 1072 👎 0


//leetcode submit region begin(Prohibit modification and deletion)

import java.util.ArrayList;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution {
    /**
     * 记录结果
     */
    List<String> res = new ArrayList<>();

    public List<String> binaryTreePaths(TreeNode root) {
        if(root == null){
            return res;
        }

        findWay(new ArrayList<>(), root);

        return res;
    }

    private void findWay(List<Integer> recordList, TreeNode cur) {
        recordList.add(cur.val);
        if (cur.left == null && cur.right == null) {
            StringBuilder s = new StringBuilder();
            for (int i = 0; i < recordList.size() - 1; i++) {
                s.append(recordList.get(i)).append("->");
            }
            s.append(recordList.get(recordList.size() - 1));
            res.add(s.toString());
            return;
        }

        if (cur.left != null) {
            findWay(recordList, cur.left);
            recordList.remove(recordList.size() - 1);
        }

        if (cur.right != null) {
            findWay(recordList, cur.right);
            recordList.remove(recordList.size() - 1);
        }

        return;
    }
}
//leetcode submit region end(Prohibit modification and deletion)