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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
If you notice carefully in the flattened tree, each node‘s right child points to the next node of a pre-order traversal.
Analyse: The sequence of the linked list is the preorder traversal. So we can first do the preorder traversal then create a new tree.
Runtime: 12ms.
1 /** 2 * Definition for a binary tree node. 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(!root) return; 14 vector<TreeNode* > vec; 15 preorder(root, vec); 16 17 TreeNode* current = root; 18 int index = 1; 19 while(index < vec.size()){ 20 TreeNode* temp(vec[index]); 21 current->right = temp; 22 current->left = NULL; 23 current = temp; 24 index++; 25 } 26 return; 27 } 28 void preorder(TreeNode* root, vector<TreeNode* >& vec){ 29 if(!root) return; 30 31 vec.push_back(root); 32 preorder(root->left, vec); 33 preorder(root->right, vec); 34 } 35 };
Flatten Binary Tree to Linked List
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原文地址：http://www.cnblogs.com/amazingzoe/p/4683466.html
