这是悦乐书的第346次更新,第370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211题(顺位题号是897).给定一棵树,按中序遍历顺序重新排列树,以便树中最左边的节点现在是树的根,并且每个节点都没有左子节点,只有一个右子节点.例如: 输入:[5,3,6,2,4,null,8,1,null,null,null,7,9] 5 / \ 3 6 / \ \ 2 4 8 / / \ 1 7 9 输出:[1,null,2,null,3,null,4,null,5,null…

897. 递增顺序查找树 897. Increasing Order Search Tree 题目描述 给定一个树,按中序遍历重新排列树,使树中最左边的结点现在是树的根,并且每个结点没有左子结点,只有一个右子结点. LeetCode897. Increasing Order Search Tree 示例: 输入: [5,3,6,2,4,null,8,1,null,null,null,7,9] ``` 5 / \ 3 6 / \ \ 2 4 8 / / \ 1 7 9 ``` 输出: [1,nul…

problem 897. Increasing Order Search Tree 参考 1. Leetcode_easy_897. Increasing Order Search Tree; 完…

题目来源: https://leetcode.com/problems/increasing-order-search-tree/ 自我感觉难度/真实难度:medium/easy 题意: 分析: 自己的代码: 优秀代码: # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class…

Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child. Example 1: Input: [5,3,6,2,4,null,8,1,null,null,null,7,9] 5 / \ 3 6 / \ \ 2 4 8  …

题目要求 Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child. 题目分析及思路 题目给出一棵树,要求重新对树进行排序,使得树中最左边的结点是树的根,并且每一个结点没有左孩子,只有一个右孩子.可以先得到已知树的中序遍历…

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 重建二叉树 数组保存节点 中序遍历时修改指针 参考资料 日期 题目地址:https://leetcode.com/problems/increasing-order-search-tree/description/ 题目描述 Given a tree, rearrange the tree in in-order so that the leftmo…

Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child. Example 1: Input: [5,3,6,2,4,null,8,1,null,null,null,7,9] 5 / \ 3 6 / \ \ 2 4 8  …

题目如下: 解题思路:我的方法是先用递归的方法找出最左边的节点,接下来再对树做一次递归中序遍历,找到最左边节点后将其设为root,其余节点依次插入即可. 代码如下: # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): n…

Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child. Example 1: Input: [5,3,6,2,4,null,8,1,null,null,null,7,9] 5 / \ 3 6 / \ \ 2 4 8  …