标签：style blog color io ar for div sp 代码
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
算法：该题用的是递归，只要考虑当前节点，思路如下，设当前节点为root，把root的左节点的最右节点的右子树指向root的右子树，然后把root的右指针指向root的左子树，root的做指针置空，下一个节点处理root->right,代码如下：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 void flatten(TreeNode *root) { 13 if(NULL==root) return; 14 if(NULL!=root->left) 15 { 16 getLeftRight(root)->right=root->right; 17 root->right=root->left; 18 root->left=NULL; 19 } 20 flatten(root->right); 21 22 } 23 24 TreeNode* getLeftRight(TreeNode* root) 25 { 26 root=root->left; 27 while(NULL!=root->right) root=root->right; 28 return root; 29 } 30 };
Flatten Binary Tree to Linked List <leetcode>
标签：style blog color io ar for div sp 代码
原文地址：http://www.cnblogs.com/sqxw/p/3970366.html
