257. Binary Tree Paths

最新推荐文章于 2024-11-03 20:00:49 发布

SnailRunFast

最新推荐文章于 2024-11-03 20:00:49 发布

阅读量155

点赞数

CC 4.0 BY-SA版权

分类专栏： Leet Code 文章标签： Binary Tree Paths

本文链接：https://blog.youkuaiyun.com/snailrunfast/article/details/70149880

Leet Code 专栏收录该内容

143 篇文章

订阅专栏

本文介绍了一种通过递归方法获取二叉树所有从根节点到叶子节点路径的方法，并提供了一个Java实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

All root-to-leaf paths are:

["1->2->5", "1->3"]

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Subscribe to see which companies asked this question.

Solution:

Tips:

tree, recursion

Java Code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<>();
        if (null == root) {
            return result;
        }
        
        binaryTreePaths(root, "", result);
        
        return result;
    }
    
    // root cann't be null
    private void binaryTreePaths(TreeNode root, String str, List<String> result) {
        if (root.left == null && root.right == null) {
            result.add(str + root.val);
            return;
        }
        
        str = str + root.val + "->";
        if (root.left != null) {
            binaryTreePaths(root.left, str, result);
        }
        
        if (root.right != null) {
            binaryTreePaths(root.right, str, result);
        }
    }
}