Flatten Binary Tree to Linked List
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
分析： 这道题目的求解思路是对于某个既含有左子树又含有右子树的节点T，寻找左子树的中最靠右的叶子节点，将T的右子树挂在该叶子节点的右节点上，再将T的左边节点移动到右边，依次遍历右子树
/** * 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) { while(root){ if(root->left && root->right){ TreeNode* t = root->left; while(t->right) t = t->right; t->right=root->right; } if(root->left){ root->right = root->left; root->left=NULL; } root = root->right; } } };
Flatten Binary Tree to Linked List
原文：http://www.cnblogs.com/willwu/p/6360434.html