LeetCode | Flatten Binary Tree to Linked List（二叉树转化成链表）

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 child points to the next node of a pre-order traversal.

题目解析：

这棵树的先序遍历是有序的。并且根据结构的特性，变形后，按照右子树链形成链表形式。

一般对树进行操作，也都用递归的形式。将做子树递归后形成的链表，插入到根结点和右子树之间。这样递归往复进行即可。

一定要有递归和模块的思想。就假设左边递归后已经形成，之后就按照链表处理就行了。这种方法比其他方法简单。

class Solution {public:    void flatten(TreeNode *root) {        if(root == NULL)            return ;        if(root->left){            flatten(root->left);            TreeNode *cur = root->left;            while(cur->right){                cur = cur->right;            }            cur->right = root->right;            root->right = root->left;            root->left = NULL;        }        if(root->right)            flatten(root->right);    }};