[LeetCode 题解]: Flatten Binary Tree to Linked List
2014-07-29 12:57
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Given a binary tree, flatten it to a linked list in-place.
For example,
Given
The flattened tree should look like:
Hints:
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
题意:将任意一颗二叉树转换成一颗仅有左子树的二叉树。
要求:(1)不得使用额外存储空间,就地排序整理。
(2)修改之后的树为仅有左子树的二叉树,节点顺序为原二叉树的先序遍历。
思路:说是结果为先序遍历,但是递归方式为后序遍历方式。
作者:Double_Win
出处: /article/5296514.html
由于本人水平有限,文章在表述和代码方面如有不妥之处,欢迎批评指正~
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.
题意:将任意一颗二叉树转换成一颗仅有左子树的二叉树。
要求:(1)不得使用额外存储空间,就地排序整理。
(2)修改之后的树为仅有左子树的二叉树,节点顺序为原二叉树的先序遍历。
思路:说是结果为先序遍历,但是递归方式为后序遍历方式。
class Solution { public: void flatten(TreeNode *root) { if(root==NULL) return; flatten(root->left); flatten(root->right); if(root->left==NULL) return; TreeNode *p = root->left; while(p->right) p=p->right; p->right=root->right; root->right=root->left; root->left=NULL; } };
作者:Double_Win
出处: /article/5296514.html
由于本人水平有限,文章在表述和代码方面如有不妥之处,欢迎批评指正~
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