Given a binary tree, return the preorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

 二叉樹先序遍歷， 根節點-> 左子樹-> 右子樹。使用一個棧來維護已經訪問過的節點。當節點不為空時，當前節點入棧，輸出節點值，繼續向左子樹遍歷。當root為空，從棧中彈出節點，向右子樹進行遍歷。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        if(root==NULL)
            return vector<int>();
        
        stack<TreeNode*> nodeStack;
        vector<int> result;
        while(!nodeStack.empty() || root != NULL)
        {
            while(root)
            {
                result.push_back(root->val);
                nodeStack.push(root);
                root=root->left;               
            }
             
            TreeNode* node = nodeStack.top();
            nodeStack.pop();  
            root = node->right;                      
        }
        return result;
        
    }
};