102. Binary Tree Level Order Traversal （java）

102. Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

题目链接：https://leetcode.com/problems/binary-tree-level-order-traversal/#/description

题目大意：中序遍历输出一棵二叉树

提交的代码：

提交的代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
//这个题目用BFS，每层把下一层的结点加进队列
public class Solution {
    
    public void BFS(TreeNode root, List<List<Integer>> ans) {    
        Queue<TreeNode> q = new LinkedList<>();          //定义队列
        q.offer(root);                                   //根值入队列
        while(!q.isEmpty()) {                         
            int size = q.size();
            List<Integer> cur = new ArrayList<>();       
            for (int i = 0; i < size; i ++) {             //遍历队列里的每一个节点
                if (q.peek().left != null) {              //队列中弹出一个节点，该节点的左子树不为空
                    q.offer(q.peek().left);               //把该节点左子树压入队列
                }
                if (q.peek().right != null) {             //同理处理右子树
                    q.offer(q.peek().right);
                }
                cur.add(q.poll().val);                    //线性表中添加该节点的val值，cur中存储的节点是在二叉树的同一层

                                           //poll：相当于先get然后再remove掉，就是查看的同时，也将这个元素从容器中删除掉
            }
            ans.add(cur);
        }

 } 
    public List<List<Integer>> levelOrder(TreeNode root) { 
       List<List<Integer>> ans = new ArrayList<>(); 
       if (root == null) { 
            return ans;   
       }
       BFS(root, ans); 
       return ans; 
    }
}

eclipse中代码：

package test_Lanqiao;

import java.util.*;

import test_Lanqiao.Main.TreeNode;

//这个题目用BFS，每层把下一层的结点加进队列
public class Main {
	//树结构体类，把public去掉
	//结构体也必须是静态的。。。。一切都要是静态的
	static class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;
		TreeNode(int x) { val = x; }
	}

    public static void BFS(TreeNode root, List<List<Integer>> ans) {
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while(!q.isEmpty()) {
            int size = q.size();
            List<Integer> cur = new ArrayList<>();
            for (int i = 0; i < size; i ++) {
                if (q.peek().left != null) {
                    q.offer(q.peek().left);
                }
                if (q.peek().right != null) {
                    q.offer(q.peek().right);
                }
                cur.add(q.poll().val);
            }
            ans.add(cur);
        }
    }
    
    public static List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        BFS(root, ans);
        return ans;
    }
    public static void main(String[] args) {
		
		   Scanner in = new Scanner(System.in);
		   Main runner = new Main();//得自己构造一棵树出来
		   TreeNode root = new TreeNode(3);
		   root.left = new TreeNode(9);
		   root.right = new TreeNode(20);
		   root.right.left = new TreeNode(15);
		   root.right.right = new TreeNode(7);
		   //树建好了，作为参数传进去就行了。但是返回值是List<List<Integer>>
		   //还得自己把结果输出
		   List<List<Integer>> ans = new ArrayList<>();
		   //因为在主函数测试，又都在同一个类，所有方法都要申明成static，这点很重要，蓝桥杯的时候就是这样，和leetcode不同，leetcode得到主函数是另外一个类。
		   ans = runner.levelOrder(root);
		   for (int i = 0; i < ans.size(); i ++) {
			   for (int j = 0; j < ans.get(i).size(); j ++) {
				   System.out.print(ans.get(i).get(j) + " ");
			   }
			   System.out.println();
		   }
	   }
}