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);
}
};