Binary Tree Preorder Traversal ---------M

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?

A:
simply use a stack to push right and left, and print the current node

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> myStack;
        if(root)
            myStack.push(root);
        while( not myStack.empty()){
            auto p = myStack.top();
            myStack.pop();
            res.push_back(p->val);
            if(p->right){
                myStack.push(p->right);
            }
            if(p->left){
                myStack.push(p->left);
            }
        }
        return res;
    }
};

Mistakes:
1:  要注意， 先往stack里加右子树，再加左子树。
2:   这次忘了先check root == nullptr