//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)
