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…
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…
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…
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 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 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…
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 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 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…