给定一个二叉树，返回它的 前序 遍历。

 示例:

输入: [1,null,2,3]

1

\

2

/

3

输出: [1,2,3]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/binary-tree-preorder-traversal

递归：

1 /** 2  * Definition for a binary tree node. 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12    13     void process(TreeNode* root,vector
     
       &ans)14     {15         ans.push_back(root->val);16         if(root->left) process(root->left,ans);17         if(root->right) process(root->right,ans);18     }19 20     vector
      
        preorderTraversal(TreeNode* root) {21         vector
       
         ans;22         if(root)23             process(root,ans);24         return ans;25     }26 27 28 };
       
      
     

迭代：

1 /** 2  * Definition for a binary tree node. 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12    13    14     vector
     
       preorderTraversal(TreeNode* root) {15         vector
      
        ans;16         if(!root) return ans;17         stack
       
         nodeStack;18         nodeStack.push(root);19         while(!nodeStack.empty()) {20             TreeNode* node = nodeStack.top();21             ans.push_back(node->val);22             nodeStack.pop();23             if(node->right) nodeStack.push(node->right);24             if(node->left) nodeStack.push(node->left);25         }26         return ans;27     }28 29 30 };