题目：
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：此题为前序遍历。 可以用迭代也可以用stack。stack的方法更直观一些。
代码：
1 class Solution { 2 public: 3 void flatten(TreeNode *root) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 if (root == NULL) return; 7 stack<TreeNode*> stk; 8 stk.push(root); 9 TreeNode np(0); 10 TreeNode* ptr = &np; 11 while(!stk.empty()) 12 { 13 TreeNode* p = stk.top(); 14 stk.pop(); 15 if(p != NULL) 16 { 17 stk.push(p->right); 18 stk.push(p->left); 19 20 ptr->right = p; 21 ptr = p; 22 ptr->left = NULL; 23 } 24 } 25 } 26 };
1 class Solution { 2 public: 3 void flatten(TreeNode *root) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 if (root == NULL) return; 7 //1.flat the left subtree 8 if (root->left) 9 flatten(root->left); 10 //2.flatten the right subtree 11 if (root->right) 12 flatten(root->right); 13 //3.if no left return 14 if (NULL == root->left) 15 return; 16 //4.insert left sub tree between root and root->right 17 //4.1.find the last node in left 18 TreeNode ** ptn = & (root->left->right); 19 while (*ptn) 20 ptn = & ((*ptn)->right); 21 //4.2.connect right sub tree after left sub tree 22 *ptn = root->right; 23 //4.3.move left sub tree to the root's right sub tree 24 root->right = root->left; 25 root->left = NULL; 26 27 } 28 };