标签：style blog color io for ar div amp
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
思路：该题本质上是二叉树的先序遍历。我们首先来看递归的解法，函数FlattenSub(node)对以node为根的树进行处理，同时返回先序序列的尾节点的指针：
1 class Solution { 2 public: 3 void flatten( TreeNode *root ) { 4 if( !root ) { return; } 5 FlattenSub( root ); 6 } 7 private: 8 TreeNode *FlattenSub( TreeNode *node ) { 9 if( node->left ) { 10 TreeNode *left_last = FlattenSub( node->left ); 11 left_last->right = node->right; 12 node->right = node->left; 13 node->left = 0; 14 return FlattenSub( left_last ); 15 } 16 if( node->right ) { return FlattenSub( node->right ); } 17 return node; 18 } 19 };
接下来再看迭代的解法，基本思想是将父节点的地址保存到其左子树最右侧节点（即先序遍历的尾节点）的右指针字段，这样在左子树处理完成之后，可以再次回到父节点以完成对右子树的处理。
1 class Solution { 2 public: 3 void flatten( TreeNode *root ) { 4 if( !root ) { return; } 5 while( root ) { 6 if( root->left ) { 7 TreeNode *node = root->left; 8 while( node->right && node->right != root ) { node = node->right; } 9 if( node->right == root ) { 10 node->right = root->right; 11 root->right = root->left; 12 root->left = 0; 13 } else { 14 node->right = root; 15 root = root->left; 16 } 17 } 18 TreeNode *node = root; 19 while( node->right && !node->right->left ) { node = node->right; } 20 if( node->right && node->right->left == root ) { 21 root = node->right; 22 node->right = root->right; 23 root->right = root->left; 24 root->left = 0; 25 } else { 26 root = node->right; 27 } 28 } 29 return; 30 } 31 };
Flatten Binary Tree to Linked List,布布扣,bubuko.com
Flatten Binary Tree to Linked List
标签：style blog color io for ar div amp
原文地址：http://www.cnblogs.com/moderate-fish/p/3916219.html
