题目：
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
2
/
3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
解析：
如果采用递归方式来做，思路是很简单的。关键是返回值result的处理：因为不能在preorderTraversal中定义vector<int>，所以定义了临时函数，把result作为一个入口参数传入其中并进行操作。
1 // recursive, but it's trivial... 2 vector<int> preorderTraversal(TreeNode* root) { 3 vector<int> v; 4 preTraversal(root, v); 5 return v; 6 } 7 void preTraversal(TreeNode* root, vector<int>& v){ 8 if(!root) return; 9 v.push_back(root->val); 10 preTraversal(root->left, v); 11 preTraversal(root->right, v); 12 }
迭代，用栈
1 vector<int> preorderTraversal(TreeNode* root) { 2 vector<int> v; 3 if(!root) return v; 4 TreeNode* temp = root; 5 stack<TreeNode*> s; 6 s.push(root); 7 while(!s.empty()){ 8 temp = s.top(); 9 s.pop(); 10 v.push_back(temp->val); 11 if(temp->right) s.push(temp->right); 12 if(temp->left) s.push(temp->left); 13 } 14 return v; 15 }
其他方法
1 vector<int> preorderTraversal(TreeNode* root) { 2 vector<int> v; 3 if(!root) return v; 4 TreeNode* temp = root; 5 stack<TreeNode*> s; 6 while(true){ 7 while(temp){ 8 v.push_back(temp->val); 9 if(temp->right) s.push(temp->right); 10 temp = temp->left; 11 } 12 if(s.empty()) break; 13 temp = s.top(); 14 s.pop(); 15 }; 16 } 17 18 // morris traversal， O(1) space 19 vector<int> preorderTraversal(TreeNode* root) { 20 vector<int> v; 21 if(!root) return v; 22 TreeNode* temp = root, *prev; 23 while(temp){ 24 if(!temp->left){ 25 v.push_back(temp->val); 26 temp = temp->right; 27 }else{ 28 prev = temp->left; 29 while(prev->right&&(prev->right != temp)) 30 prev = prev->right; 31 if(!prev->right){ 32 v.push_back(temp->val); 33 prev->right = temp; 34 temp = temp->left; 35 }else{ 36 prev->right = NULL; 37 temp = temp->right; 38 } 39 } 40 } 41 }