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 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…

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 t…

114. Flatten Binary Tree to Linked List (Medium) 453. Flatten Binary Tree to Linked List (Easy) 解法1: 用stack. class Solution { public: void flatten(TreeNode *root) { if(!root) return; TreeNode *pnode = NULL; stack<TreeNode*> s; s.push(root); while(!s…

. 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…

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…

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 SOLUTION 1:使用递归解决,根据left是否为空,先连接left tree, 然后再连接右子树.使用一个t…

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 这题的主要难点是"in-place".因为这个flatten的顺序是中.左.右(相当于中序遍历),所…

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 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…

1.  Path Sum Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example: Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \ 7…

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 My Submissions QuestionEditorial Solution Total Accepted: 81373 Total Submissions: 261933 Difficulty: Medium Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The fl…

Problem Link: http://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/ The problem is asking for flatterning a binary tree to linked list by the pre-order, therefore we could flatten tree from the root. For each node, we link it with its n…

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 思路: 按照树的先序遍历顺序把节点串联起来即可. /** * Definition for a binary tree node. * function TreeNode(va…

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 Accepted 218,918 Submissions 533,947 [解析]由上图可知,如果右子树不为空,则右子树最后肯定为左子树最有一…

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…

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.…

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. For example,Given 1 / \ 2 5 / \ \ 3 4 6   The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6…

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 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 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 题目大意:给一个二叉树,将它转为一个list,转换后的序列应该是先序遍历,list由这棵树的右孩子表示. 解题思路:递归的处理左子树,然后处理右子树,将左孩子的右孩子置为当前节点的右孩…

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: 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 题解:如hint所给出,这道题就是使用先序遍历,遍历到的值作为新的右孩子存起来,左孩子变为空.注意的是,因为右孩子会更新,所以为了递归右子树,要在更新之前提前保存右孩子.整个…

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 the…

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: 看上去很简单,使用二叉树的中序遍历就可以实现.但是题目的小曲点在于要在原tree上修改,如果将左儿子放在右儿子位置上,会丢失右子树,导致遍历失败. 解决方法有两种: 1.不…

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 the…

本质上是二叉树的root->right->left遍历. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 给定一个二叉树,就地的把他转换成一个链表. 例如: 给定 1 / \ 2 5 / \ \ 3 4 6 转换后的树应该向这样子: 1 \ 2 \ 3 \ 4 \ 5 \ 6 提示: 如果你观察的足够仔细,你会发现,每个还在的右孩子指针指向了,这个节点在前序遍历中的后面的那个节…

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…