Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 题目 将左子树所形成的链表插入到root和root->right之间 思路 1 / 2(root) 假设当前root为2 / \ 3(p) 4…

Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 给一个二叉树,把它展平为链表 in-place 根据展平后的链表的顺序可以看出是先序遍历的结果,所以用inorder traversal. 解…

题目 给定一个二叉树,原地将它展开为链表. 例如,给定二叉树 1 / \ 2 5 / \ \ 3 4 6 将其展开为: 1 \ 2 \ 3 \ 4 \ 5 \ 6 解析 通过递归实现:可以用先序遍历,然后串成链表 主要思想就是:先递归对右子树进行链表化并记录,然后将root->right指向 左子树进行链表化后的头结点,然后一直向右遍历子树,连接上之前的右子树 /** * Definition for a binary tree node. * struct TreeNode { * int v…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 click to show hints. Hints: If you notice carefully in the flattened tree, each node's right…

/* 先序遍历构建链表,重新构建树 */ LinkedList<Integer> list = new LinkedList<>(); public void flatten(TreeNode root) { preOrder(root); TreeNode res = root; list.poll(); while (!list.isEmpty()) { res.right = new TreeNode(list.poll()); res.left = null; res =…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 这道题就是将数变为只有右节点的树. 递归,不是很难. 递归的时候求出左节点的最右孩子即可. /** * Definition for a binary tree node. * pub…

Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 这个题思路就是DFS, 先左后右, 记住如果是用stack如果需要先得到左, 那么要先append右, 另外需要注意的就是用recursive…

Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6解题思路:试图通过排序后new TreeNode是无法通过的,这道题的意思是把现有的树进行剪枝操作.JAVA实现如下: static public void flatten(TreeN…

Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 思路:这题主要是就是前序遍历.主要解法就是将左子树转换为右支树,同一时候加入在右子树前.左子树求解时,须要主要左子树的深度. 详细代码例如以下: /** * Definition f…

［抄题］: Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 [暴力解法]: 时间分析: 空间分析: [优化后]: 时间分析: 空间分析: ［奇葩输出条件］: ［奇葩corner case］:…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6Hints: If you notice carefully in the flattened tree, each node's right child points to the n…

根据提示,本题等价于pre order traverse遍历,并且依次把所有的节点都存成right child,并把left child定义成空集.用递归的思想,那么如果分别把左右子树flatten成list,我们有: 1 /    \ 2     5 \       \ 3      6 <- rightTail \ 4  <- leftTail 所以使用递归的解法一: 注意由于右子树最后要接到左子树的后面,所以用temp保存右子树的head. def flatten(self, root)…

把二叉树先序遍历,变成一个链表,链表的next指针用right代替 用递归的办法先序遍历,递归函数要返回子树变成链表之后的最后一个元素 class Solution { public: void helper(TreeNode* cur, TreeNode*& tail){ //a(tail) //lk("root",tail) //a(cur) //lk("root",cur) //dsp tail=cur; TreeNode* right=cur->…

114 Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place. 将二叉树展开成链表 [] (D:\dataStructure\Leetcode\114.png) 思路:将根节点与左子树相连,再与右子树相连.递归地在每个节点的左右孩子节点上,分别进行这样的操作. 代码 class Solution(object): def flatten(self, root): i…

Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 click to show hints. Hints: If you notice carefully in th…

. Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: / \ / \ \ The flattened tree should look like: \ \ \ \ \ /** * Definition for a binary tree node. * struct TreeNode…

随笔一记,留做重温! Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 第一个想法是先序遍历,然后按照访问顺序,添加右结点. public static void…

Flatten Binary Tree to Linked List Total Accepted: 25034 Total Submissions: 88947My Submissions Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5…

Flatten a binary tree to a fake "linked list" in pre-order traversal. Here we use the right pointer in TreeNode as the next pointer in ListNode. Notice Don't forget to mark the left child of each node to null. Or you will get Time Limit Exceeded…

题目来源 https://leetcode.com/problems/flatten-binary-tree-to-linked-list/ Given a binary tree, flatten it to a linked list in-place. 题意分析 Input: binary tree Output: flattened tree Conditions:将一个二叉树压平为一个flatten 树,也就是一条斜线 题目思路 先将左右子树压平,然后将左子树嵌入到本节点与右子树之间.…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \…

题目说明 Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 题目分析 第一感觉是前序遍历,顺便打算在这题练习一下昨天学到的二级指针的写法XD,调的时候bug挺多的,可读性贼差,指针还是慎用啊-- 以下为个人实现(C++,12ms):…

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 先序遍历 递归 日期 题目地址:https://leetcode.com/problems/flatten-binary-tree-to-linked-list/#/description 题目描述 Given a binary tree, flatten it to a linked list in-place. For example, Give…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 click to show hints. Subscribe to see which companies asked this question 解答 先序遍历同时把节点都堆到左边,…

题目 Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 解题思路:利用递归找到倒数第一个父节点.记录下它的右节点,将左边的移到右边.然后再把之前标记的右节点连接上. 代码 public class Solution { public…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 click to show hints. Hints: If you notice carefully in the flattened tree, each node's right…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6   The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 Hints: If you notice carefully in the flattened tree, each node's right child points to th…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 思路: 用先序遍历,得到的是从小到大的顺序.把先序遍历变形一下: void flatten(TreeNode* root) { if(NULL == root) return; vec…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6代码如下: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNod…

Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 法I:递归,前序遍历 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNo…