leetcode第47题（binary-tree-level-order-traversal）

二叉树层次遍历

最新推荐文章于 2021-04-27 21:46:40 发布

原创最新推荐文章于 2021-04-27 21:46:40 发布 · 305 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#递归 #队列

牛客网（leetcode）专栏收录该内容

47 篇文章

订阅专栏

本文介绍了一种二叉树的层次遍历算法，并通过Java实现。该算法按从左到右、由上至下的顺序返回节点的值，每层为一个列表。

题目：

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,#,#,15,7},

return its level order traversal as:

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

confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

思路：

跟前面42题一样，只是最后加的时候不加在头部即可。

代码：

import java.util.ArrayList;
import java.util.Collections;
import java.util.LinkedList;
import java.util.Queue;
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(root==null){
            return result;
        }
        Queue<TreeNode> que = new LinkedList<TreeNode>();
        que.add(root);
        while(!que.isEmpty()){
            int count = que.size();
            ArrayList<Integer> list = new ArrayList<Integer>();
            for(int i=0;i<count;i++){
                TreeNode node = que.poll();
                list.add(node.val);
                if(node.left!=null){
                    que.add(node.left);
                }
                if(node.right != null){
                    que.add(node.right);
                }
            }
            result.add(list);
        }
        return result;
    }
}