Given an n-ary tree, return the preorder traversal of its nodes' values.
For example, given a 3-ary tree:
Return its preorder traversal as: [1,3,5,6,2,4].
Note:
Recursive solution is trivial, could you do it iteratively?
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额，这个迭代不会，不过，如果会了二叉树的前序遍历的递归解法的话，解这个题就会感觉简单了，同理，后序遍历也是，不过，中序遍历上可能会很难写。
C++代码：
/*
// Definition for a Node.
class Node {
public:
int val;
vector<Node*> children; Node() {} Node(int _val, vector<Node*> _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public:
vector<int> preorder(Node* root) {
vector<int> res;
helper(root,res);
return res;
}
void helper(Node *root,vector<int> &res){
if(!root) return;
res.push_back(root->val);
for(Node* cur : root->children){
helper(cur,res);
}
}
};