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
思路: 首先找到根节点左节点的最右子节点,然后把根节点的右子树移到左子树的最右端;接着把根结点的左子树移到右节点上,并将左子树置为空树,同时将根结点往右下移一个节点,依次递归即可。代码如下:
Code(c++):
/**
* 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 flatten(TreeNode* root) {
if(!root) return;
while(root){
if(root->left && root->right){
TreeNode *p = root->left;
while(p->right)
p =p->right;
p->right = root->right;
}
if(root->left)
root->right = root->left;
root->left = NULL;
root = root->right;
}
}
};
Leetcode[114]-Flatten Binary Tree to Linked List
原文:http://blog.csdn.net/dream_angel_z/article/details/46481999