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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
此题看题目给出的示例就知道是先序遍历
本题用递归求先序遍历
class Solution { public: vector<TreeNode *> res; void preTraverse(TreeNode *root){ if(root == NULL) return; res.push_back(root); preTraverse(root->left); preTraverse(root->right); } void flatten(TreeNode *root) { if(root == NULL) return; preTraverse(root); for(int i = 1 ; i < res.size(); ++ i){ res[i-1]->left =NULL; res[i-1]->right = res[i]; } res[res.size()-1]->left =NULL; res[res.size()-1]->right = NULL; } };
下面用迭代求解，一步步的构造结点，由于题目是先序遍历，可以将右子树链接到左子树最右的孩子结点，如果不明白的可以手动模拟一下看与先序遍历是否相同,然后从根结点一步步往下构造
class Solution { public: void flatten(TreeNode *root) { if(root == NULL) return; while(root){ if(root->left){ TreeNode* ptr = root->left; while(ptr->right) ptr = ptr->right; ptr->right = root->right; root->right = root->left; root->left = NULL; } root = root->right; } } };
Leetcode Flatten Binary Tree to Linked List,布布扣,bubuko.com
Leetcode Flatten Binary Tree to Linked List
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原文地址：http://www.cnblogs.com/xiongqiangcs/p/3821054.html
