144. Binary Tree Preorder Traversal

iven 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].

solution:

/** * 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) return {};        vector<int> res;        stack<TreeNode*> stk;        stk.push(root);        while(!stk.empty()){            TreeNode *tmp = stk.top();            res.push_back(tmp->val);            stk.pop();            if(tmp->right) stk.push(tmp->right);            if(tmp->left)  stk.push(tmp->left);        }        return res;    } };

思路：用栈的地方基本可以用递归， 主要考虑的是循环的终止时间

运行速度：快