[Leetcode] Flatten Binary Tree to Linked List

题目链接在此

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

就是把一棵树按先序铺展成一颗斜向右的树。

如果先序遍历一边把值都记下来再建树，那自然是简单。可是怎样用常数空间O（1）来解决呢？

转自这位大神

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
void flatten(TreeNode* root) {
TreeNode* now = root;
while (now) {
if (now->left) {
TreeNode* pre = now->left;
//  找now的左子树的最右子树
while (pre->right) {
pre = pre->right;
}
//把now的右子树，作为【左子树的最右子树】的右子树，
pre->right = now->right;
//把now的左子树，作为now的右子树
now->right = now->left;
now->left = NULL;
}
now = now->right;
}
}
};