N-ary Tree Preorder Traversal

这题的数据结构还是上一题的，很简单，熟悉二叉树遍历的看一眼就写出来了。先看题目：

Given an n-ary tree, return the preorder traversal of its nodes' values.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

 

Follow up:

Recursive solution is trivial, could you do it iteratively?

 

Example 1:



Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]
Example 2:



Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]
 

Constraints:

The height of the n-ary tree is less than or equal to 1000
The total number of nodes is between [0, 10^4]

先序遍历：

class Solution {
public:
    void visit(Node *node, vector<int> &vec) {
        // Visit the current node
        vec.push_back(node->val);
        if (node->children.empty()) {
            return;
        }
        for (auto v : node->children) {
            visit(v, vec);
        }
    }

    vector<int> preorder(Node* root) {
        vector<int> res;
        if (nullptr == root) {
            return res;
        }
        visit(root, res);
        return res;
    }
};