Formatted question description: https://leetcode.ca/all/589.html

589. N-ary Tree Preorder Traversal (Easy)

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?

Solution 1. Recursive

// OJ: https://leetcode.com/problems/n-ary-tree-preorder-traversal/

// Time: O(N)
// Space: O(logN)
class Solution {
private:
    vector<int> ans;
    void rec(Node *root) {
        if (!root) return;
        ans.push_back(root->val);
        for (auto ch : root->children) rec(ch);
    }
public:
    vector<int> preorder(Node* root) {
        rec(root);
        return ans;
    }
};

Solution 2. Iterative

// OJ: https://leetcode.com/problems/n-ary-tree-preorder-traversal/

// Time: O(N)
// Space: O(logN)
class Solution {
public:
    vector<int> preorder(Node* root) {
        if (!root) return {};
        vector<int> ans;
        stack<Node*> s;
        s.push(root);
        while (s.size()) {
            root = s.top();
            s.pop();
            ans.push_back(root->val);
            for (int i = root->children.size() - 1; i >= 0; --i) {
                if (root->children[i]) s.push(root->children[i]);
            }
        }
        return ans;
    }
};

Java

  • /*
    // Definition for a Node.
    class Node {
        public int val;
        public List<Node> children;
    
        public Node() {}
    
        public Node(int _val) {
            val = _val;
        }
    
        public Node(int _val, List<Node> _children) {
            val = _val;
            children = _children;
        }
    };
    */
    class Solution {
        public List<Integer> preorder(Node root) {
            List<Integer> preorder = new ArrayList<Integer>();
            if (root == null)
                return preorder;
            Stack<Node> stack = new Stack<Node>();
            stack.push(root);
            while (!stack.isEmpty()) {
                Node node = stack.pop();
                preorder.add(node.val);
                List<Node> children = node.children;
                int numOfChildren = children.size();
                for (int i = numOfChildren - 1; i >= 0; i--)
                    stack.push(children.get(i));
            }
            return preorder;
        }
    }
    
  • // OJ: https://leetcode.com/problems/n-ary-tree-preorder-traversal/
    // Time: O(N)
    // Space: O(logN)
    class Solution {
    private:
        vector<int> ans;
        void rec(Node *root) {
            if (!root) return;
            ans.push_back(root->val);
            for (auto ch : root->children) rec(ch);
        }
    public:
        vector<int> preorder(Node* root) {
            rec(root);
            return ans;
        }
    };
    
  • """
    # Definition for a Node.
    class Node(object):
        def __init__(self, val, children):
            self.val = val
            self.children = children
    """
    class Solution(object):
        def preorder(self, root):
            """
            :type root: Node
            :rtype: List[int]
            """
            res = []
            if not root:
                return res
            res.append(root.val)
            for child in root.children:
                res.extend(self.preorder(child))
            return res
    

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