114. flatten binary tree to linked list

每日一题 2019 - 04 - 27

题目:

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

For example, given the following tree:

1
2
3
4
5
    1
/
2 5
/
3 4 6

The flattened tree should look like:

1
2
3
4
5
6
7
8
9
10
11
1

2

3

4

5

6

解法:

这个题让把给出的二叉树转成一个链表,这个链表特征为:所有二叉树的结点链接在二叉树的右子树上,解题思路如下:

  • 深层次遍历二叉树,找到最下层的左子节点
  • 将左子分支加入到右子分支中,也即
1
2
3
4
5
6
7
  1
/
2 5

3 6

4
  • 重复上述过程,直到全部归并到右子分支中;

代码:

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* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class {
public:
void flatten(TreeNode* root) {
if(root == NULL)
{
return ;
}
if(root->left)
{
flatten(root->left);
}
if(root->right)
{
flatten(root->right);
}
TreeNode* temp = root -> right ;
root -> right = root -> left ;
root -> left = NULL ;
while(root->right)
{
root = root -> right ;
}
root -> right = temp ;
}
};