---
tags: leetcode
---

# [589. N-ary Tree Preorder Traversal](https://leetcode.com/problems/n-ary-tree-preorder-traversal/)
---
# My Solution
## Solution 1: recursive algorithm
### The Key Idea for Solving This Coding Question
### C++ Code

/*
// 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> preorder(Node *root) {
        vector<int> answer;

        dfs(root, answer);
        return answer;
    }

private:
    void dfs(Node *root, vector<int> &answer) {
        if (root == nullptr) {
            return;
        }

        answer.push_back(root->val);
        for (Node *next : root->children) {
            dfs(next, answer);
        }
    }
};

### Time Complexity
$O(n)$
$n$ is the number of nodes in the n-ary tree referred by `root`.
### Space Complexity
$O(H)$
$H$ is the height the n-ary tree referred by `root`.
## Solution 2: Iteration and stack
### The Key Idea for Solving This Coding Question
### C++ Code

/*
// 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> preorder(Node *root) {
        if (root == nullptr) {
            return {};
        }
        
        vector<int> answer;
        stack<Node *> st;
        
        st.push(root);
        while (!st.empty()) {
            Node *node = st.top();
            st.pop();
            
            answer.push_back(node->val);
            for (int i = node->children.size() - 1; i >= 0; --i) {
                if (node->children[i] != nullptr) {
                    st.push(node->children[i]);
                }
            }
        }
        
        return answer;
    }
};

### Time Complexity
$O(n)$
$n$ is the number of nodes in the n-ary tree referred by `root`.
### Space Complexity
$O(H)$
$H$ is the height the n-ary tree referred by `root`.
# Miscellane
<!--
# Test Cases

[1,null,3,2,4,null,5,6]

[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]

[]

[1]

-->