Flatten Binary Tree to Linked List问题及解法

问题描述：

Given a binary tree, flatten it to a linked list in-place.

示例：

Given

1
/ \
2   5
/ \   \
3   4   6

The flattened tree should look like:

1
\
2
\
3
\
4
\
5
\
6

问题分析：
由题意可知，我们可以先把树的结点按照先序遍历存到数组vec中，然后遍历数组，修改指针，首尾相接即可。

过程详见代码：

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void preorder(TreeNode* root, vector<TreeNode*>& vec){
if(root == NULL)
return;
vec.push_back(root);
preorder(root->left, vec);
preorder(root->right, vec);
}

void flatten(TreeNode* root) {
vector<TreeNode*> vec;
preorder(root, vec);

for(int i = 0; i < (int)vec.size()-1; ++i){
vec[i]->right = vec[i+1];
vec[i]->left = NULL;
}

}
};