乐趣区

关于leetcode:leetcode-257-Binary-Tree-Paths-二叉树的所有路径简单

一、题目粗心

标签: 搜寻

https://leetcode.cn/problems/binary-tree-paths

给你一个二叉树的根节点 root,按 任意程序,返回所有从根节点到叶子节点的门路。

叶子节点 是指没有子节点的节点。

示例 1:

输出:root = [1,2,3,null,5]
输入:[“1->2->5″,”1->3”]

示例 2:

输出:root = [1]
输入:[“1”]

提醒:

树中节点的数目在范畴 [1, 100] 内
-100 <= Node.val <= 100

二、解题思路

dfs 解决即可

三、解题办法

3.1 Java 实现

/**
 * 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 {public List<String> binaryTreePaths(TreeNode root) {List<String> ans = new ArrayList<>();
        dfs(root, "", ans);
        return ans;
    }

    void dfs(TreeNode node, String path, List<String> ans) {
        boolean end = true;
        if ("".equals(path)) {path = String.valueOf(node.val);
        } else {path += "->" + node.val;}
        if (node.left != null) {dfs(node.left, path, ans);
            end = false;
        }
        if (node.right != null) {dfs(node.right, path, ans);
            end = false;
        }
        if (end) {ans.add(path);
        }
    }
}

四、总结小记

  • 2022/6/11 做完前几个中等的搜寻题,再做简略的就很简略了
退出移动版