题解 | #二叉树层序遍历#

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param root TreeNode类
     * @return int整型二维数组
     */
    public int[][] levelOrder (TreeNode root) {
        Queue<TreeNode> treeNodeQueue = new LinkedList<>();
        if (root != null) {
            treeNodeQueue.add(root);
        }
        ArrayList<ArrayList<Integer>> arrayLists = new ArrayList<>();
        while (!treeNodeQueue.isEmpty()) {
            int size = treeNodeQueue.size();
            ArrayList<Integer> tempArrayList = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode treeNode = treeNodeQueue.poll();
                assert treeNode != null;
                tempArrayList.add(treeNode.val);
                if (treeNode.left != null) {
                    treeNodeQueue.add(treeNode.left);
                }
                if (treeNode.right != null) {
                    treeNodeQueue.add(treeNode.right);
                }
            }
            arrayLists.add(tempArrayList);
        }
        int [][] result = new int[arrayLists.size()][];
        for (int i = 0; i < arrayLists.size(); i++) {
            ArrayList<Integer> arrayList = arrayLists.get(i);
            int [] temp = new int[arrayList.size()];
            for (int j = 0; j < arrayList.size(); j++) {
                temp[j] = arrayList.get(j);
            }
            result[i] = temp;
        }
        return result;
    }
}

本题知识点分析：

1.二叉树

2.队列

3.有序集合

4.集合转数组

本题解题思路分析：

1.经典的层序遍历，注意根节点是否为空即可

2.注意集合转二维数组怎么转化

3.防止空指针，我这边用断言作了判断，用if也可以

本题使用编程语言: Java