Leetcode 589. N-ary Tree Preorder Traversal

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【LeetCode】144. Binary Tree Preorder Traversal (3 solutions)

Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 解法一:递归 /** *…

【LeetCode】144. Binary Tree Preorder Traversal 解题报告（Python&C++&Java）

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录题目描述题目大意解题方法递归迭代日期题目地址:https://leetcode.com/problems/binary-tree-preorder-traversal/#/description 题目描述 Given a binary tree, return the preorder traversal of its nodes' values. For exampl…

【LeetCode】144. Binary Tree Preorder Traversal

题目: Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 提示: 首先需要明确前序遍历的顺序,即:节点 -> 左孩子 -> 右孩子,这一…

LeetCode OJ 144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? Subscribe to see which companies asked this…

LeetCode OJ：Binary Tree Preorder Traversal（前序遍历二叉树）

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3} 1 \ 2 / 3 return [1,2,3]. 前序遍历二叉树,只不过题目的要求是尽量不要使用递归,当然我还是先用递归写了一个: /** * Definition for a binary tree node. * struct TreeNode { * int val…

LeetCode算法题-N-ary Tree Preorder Traversal（Java实现）

这是悦乐书的第268次更新,第282篇原创 01 看题和准备今天介绍的是LeetCode算法题中Easy级别的第135题(顺位题号是589).给定一个n-ary树,返回其节点值的前序遍历.例如,给定一个3-ary树: 1 / | \ 3 2 4 / \ 5 6 其前序遍历结果为:[1,3,5,6,2,4]. 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试. 02 第一种解法利用递归.二叉树前序遍历是从根节点开始,然后是…

【LEETCODE OJ】Binary Tree Preorder Traversal

Problem Link: http://oj.leetcode.com/problems/binary-tree-preorder-traversal/ Even iterative solution is easy, just use a stack storing the nodes not visited. Each iteration, pop a node and visited it, then push its right child and then left child in…