LeetCode 589. N-ary Tree Preorder Traversal

原题链接在这里：https://leetcode.com/problems/n-ary-tree-preorder-traversal/

题目：

Given 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).

Follow up:

Recursive solution is trivial, could you do it iteratively?

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 height of the n-ary tree is less than or equal to 1000
• The total number of nodes is between [0, 10^4]

题解：

Recursion is obvious.

Time Compelxity: O(V+E).

Space: O(V). stack space.

AC 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> res = new ArrayList<Integer>();
        if(root == null){
            return res;
        }
        
        dfs(root, res);
        return res;
    }
    
    private void dfs(Node root, List<Integer> res){
        if(root == null){
            return;
        }
        
        res.add(root.val);
        for(Node child : root.children){
            dfs(child, res);
        }
    }
}

Iteration method, add root into stack.

When popping up top node, for all its children, iterate from right to left. Add each of them to stack.

Time Complexity: O(V+E).

Space: O(V).

AC 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> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        
        Stack<Node> stk = new Stack<>();
        stk.push(root);
        while(!stk.isEmpty()){
            Node cur = stk.pop();
            res.add(cur.val);
            List<Node> nexts = cur.children;
            if(nexts != null){
                for(int i = nexts.size()-1; i>=0; i--){
                    stk.push(nexts.get(i));
                }
            }
        }
        
        return res;
    }
}

类似Binary Tree Preorder Traversal, N-ary Tree Postorder Traversal.