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
Hints:
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
二叉树转换,转换后的形式中节点顺序恰好是二叉树前序遍历后的顺序。
解法,对二叉树的左子树作转换操作,对二叉树的右子树作转换操作,然后将左子树转换结果拼接到根节点和右子树转换结果中间,执行过程如下图所示。
1 / \ 2 5 / \ \ 3 4 6
1
/ \
2 5
\ \
3 6
\
4
1 \ 2 \ 3 \ 4 \ 5 \ 6AC code:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
public:
void flatten(TreeNode *root)
{
if(root==NULL)
return;
flatten(root->left);
flatten(root->right);
if(root->left!=NULL)
{
TreeNode *p=root->left;
while(p->right!=NULL)
p=p->right;
p->right=root->right;
root->right=root->left;
root->left=NULL;
}
}
};