LeetCode--589--easy--NaryTreePreorderTraversal

package com.app.main.LeetCode.tree;

import com.app.main.LeetCode.prebase.Node;

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

/**
 * 589
 *
 * easy
 *
 * https://leetcode.com/problems/n-ary-tree-preorder-traversal/
 *
 * 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]
 *
 *
 * Created with IDEA
 * author:Dingsheng Huang
 * Date:2019/12/6
 * Time:下午4:28
 */
public class NaryTreePreorderTraversal {

    public List<Integer> preorder(Node root) {
        List<Integer> result = new ArrayList<>();
        preorderTraversal(root, result);
        return result;
    }

    private void preorderTraversal(Node curr,  List<Integer> result) {
        if (curr != null) {
            result.add(curr.val);
            if (curr.children != null) {
                for (int i = 0; i < curr.children.size(); i++) {
                    preorderTraversal(curr.children.get(i), result);
                }
            }
        }
    }

    public List<Integer> preorder2(Node root) {
        List<Integer> result = new ArrayList<>();
        Stack<Node> stack = new Stack<>();
        Node curr = root;
        stack.add(curr);
        while (!stack.isEmpty()) {
            curr = stack.pop();
            result.add(curr.val);
            for (int i = curr.children.size() - 1; i >= 0; i--) {
                stack.push(curr.children.get(i));
            }
        }
        return  result;

    }
}
来源：CSDN
作者：微观尽头
链接：https://blog.csdn.net/huangdingsheng/article/details/103464947
标签
leetcode