589

589. N-ary Tree Preorder Traversal (Easy)

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.

 

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

Companies: Google

Related Topics:
Stack, Tree, Depth-First Search

Similar Questions:

• Binary Tree Preorder Traversal (Easy)
• N-ary Tree Level Order Traversal (Medium)
• N-ary Tree Postorder Traversal (Easy)

Solution 1. Recursive

// OJ: https://leetcode.com/problems/n-ary-tree-preorder-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    vector<int> preorder(Node* root) {
        vector<int> ans;
        function<void(Node*)> dfs = [&](Node *root) {
            if (!root) return;
            ans.push_back(root->val);
            for (auto n : root->children) dfs(n);
        };
        dfs(root);
        return ans;
    }
};

Solution 2. Iterative

// OJ: https://leetcode.com/problems/n-ary-tree-preorder-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> preorder(Node* root) {
        if (!root) return {};
        vector<int> ans;
        stack<Node*> s{{root}};
        while (s.size()) {
            root = s.top();
            s.pop();
            ans.push_back(root->val);
            for (int i = (int)root->children.size() - 1; i >= 0; --i) s.push(root->children[i]);
        }
        return ans;
    }
};