LeetCode 144:Binary Tree Preorder Traversal

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?

 //递归方法：利用一个中间函数，将res作为输入参数 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> res; return tmpFunction(root, res); } vector<int> tmpFunction(TreeNode* root, vector<int> &res){ if (root == NULL) return res; res.push_back(root->val); tmpFunction(root->left,  res); tmpFunction(root->right, res); return res; } };

 //非递归方法： //利用一个栈结构,现将头结点root压入栈中，然后从stack中弹出栈顶节点，记为tmp，然后将tmp的值压入vector<int> res中， //再将节点tmp的右孩子（不为空的话）先压入stack中，最后将tmp的左孩子（不为空的话）压入stack中， //不断重复以上操作，直到stack为空 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> res; stack<TreeNode*> s; TreeNode* tmp; if (root == NULL) return res; else { s.push(root); while (!s.empty()) { tmp = s.top(); s.pop(); res.push_back(tmp->val); if (tmp->right != NULL)  s.push(tmp->right); if (tmp->left != NULL)    s.push(tmp->left); } } return res; } };