Leetcode 257. Binary Tree Paths

最新推荐文章于 2025-01-12 23:00:00 发布

转载最新推荐文章于 2025-01-12 23:00:00 发布 · 97 阅读

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原文链接：http://www.cnblogs.com/xiagnming/p/9602637.html

本文详细解析了LeetCode上的一道经典二叉树题目——寻找所有从根节点到叶子节点的路径。通过递归算法，文章展示了如何高效地解决这个问题，并提供了完整的Java代码实现。

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题目链接

https://leetcode.com/problems/binary-tree-paths/description/

题目描述

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

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

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

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

题解

采用递归遍历，每次先判断是否到达根节点，如果到达根节点，就添加到List里面，然后清空新添加的字符，回到递归前的状态

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> list = new ArrayList<>();
        if (root == null) {
            return list;
        }
        StringBuilder sb = new StringBuilder();
        findHelper(root, list, sb);
        return list;
    }
    public void findHelper(TreeNode root, List<String> list, StringBuilder sb) {
        if (root == null) { return ; }
        
        int len = sb.length();
        if (root.left == null && root.right == null) {
            sb.append(root.val);
            list.add(sb.toString());
            sb.delete(len, sb.length());
            return ;
        }
        sb.append(root.val).append("->");
        findHelper(root.left, list, sb);
        findHelper(root.right, list, sb);
        sb.delete(len, sb.length());
    }
}

转载于:https://www.cnblogs.com/xiagnming/p/9602637.html