一、题目
Given the root of 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)
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 number of nodes in the tree is in the range [0, 104].
0 <= Node.val <= 104
The height of the n-ary tree is less than or equal to 1000.文章来源：https://www.toymoban.com/news/detail-826152.html
Follow up: Recursive solution is trivial, could you do it iteratively?文章来源地址https://www.toymoban.com/news/detail-826152.html
二、题解
/*
// Definition for a Node.
class Node {
public:
int val;
vector<Node*> children;
Node() {}
Node(int _val) {
val = _val;
}
Node(int _val, vector<Node*> _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public:
vector<int> res;
void npreorder(Node* root){
if(!root) return;
res.push_back(root->val);
for(int i = 0;i < root->children.size();i++){
npreorder(root->children[i]);
}
}
vector<int> preorder(Node* root) {
npreorder(root);
return res;
}
};
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