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, 10^4].
0 <= Node.val <= 10^4
The height of the n-ary tree is less than or equal to 1000.

Follow up: Recursive solution is trivial, could you do it iteratively?

Solution

Recursion

Time complexity : O(n)
Space complexity : O(n)

/*
// 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) {
        ans = {};
        dfs(root);
        return ans;
    }
private:
    vector<int> ans;
    void dfs(Node *node) {
        if (node == nullptr) return;
        ans.push_back(node->val);
        for (Node *c: node->children) {
            if (c == nullptr) continue;
            dfs(c);
        }
    }
};

Iteration

Time complexity : O(n) Space complexity : O(n)

/*
// 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> ans = {};
        stack<Node*> stk;
        stk.push(root);
        while (!stk.empty()) {
            Node *curr = stk.top(); stk.pop();
            if (curr == nullptr) continue;
            ans.push_back(curr->val);
            for (int i=curr->children.size()-1; 0<=i; --i)
                stk.push(curr->children[i]);
        }
        return ans;
    }
};