Problem Description:
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
If you notice carefully in the flattened tree, each node‘s right child points to the next node of a pre-order traversal.
/**
* 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;
while (root != NULL)
{
if (root->left != NULL)
{
TreeNode *p = root->left;
while (p->right != NULL) <span style="font-family: Arial, Helvetica, sans-serif;">//找到左孩子的最右边节点</span>
{
p = p->right;
}
p->right = root->right;
root->right = root->left;
root->left = NULL;
}
root = root->right;
}
}
};/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *pre=NULL;
void flatten(TreeNode *root) {
if(root==NULL)
return;
TreeNode *lastright=root->right;//记录当前节点的右子树
if(pre)
{
pre->left=NULL;
pre->right=root;
}
pre=root;
flatten(root->left);
flatten(lastright);
}
};Flatten Binary Tree to Linked List
原文：http://blog.csdn.net/longhopefor/article/details/39084259