637. Average of Levels in Binary Tree（一棵树每层节点的平均数）（二叉树的层序遍历）

Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

Example 1:

Input:
    3
   / 
  9  20
    /  
   15   7
Output: [3, 14.5, 11]
Explanation:
The average value of nodes on level 0 is 3,  on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].

Note:

1. The range of node's value is in the range of 32-bit signed integer.

这个题主要想总结一下bfs算法：

广度优先遍历：类似于树的层序遍历。将根放入队列，一层层遍历，将根弹出后，将左右孩子加入队列。

1、树的广度优先遍历：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public ArrayList<Integer> bfs(TreeNode root){
        ArrayList<Integer> list=new ArrayList<Integer>();
        if(root==null) return list;
        Queue<TreeNode> queue=new LinkedList<TreeNode>();
        queue.add(root);
        while (!queue.isEmpty()){
            TreeNode treeNode=queue.poll();
            if(treeNode.left!=null)  queue.add(treeNode.left);
            if(treeNode.right!=null)  queue.add(treeNode.right);
            list.add(treeNode.val);
        }
        return list;
    }
}

2、图的广度优先遍历

 以s为顶点的邻接表为：

　　S -> A -> B

　　A -> C -> D

　　B -> E

class Solution {
    static HashMap<Character,LinkedList<Character>> graph;//输入的邻接表
    static HashMap<Character,Integer> dirt;//距离集合
    public static void bfs(HashMap<Character,LinkedList<Character>> graph,
                           HashMap<Character,Integer> dirt,char start){
        Queue<Character> queue=new LinkedList<Character>();
        queue.add(start);//把起点加入队列
        dirt.put(start,0);//记录每个节点距离顶点的距离
        while (!queue.isEmpty()){
            char top=queue.poll();//取出队列顶的元素
            int d=dirt.get(top)+1;
            for(Character c:graph.get(top)){
                if(!dirt.containsKey(c)){
                    dirt.put(c,d);
                    queue.add(c);
                }
            }
        }
    }
    public static void main(String[] args) {
        // s顶点的邻接表
        LinkedList<Character> list_s = new LinkedList<Character>();
        list_s.add('A');
        list_s.add('B');
        LinkedList<Character> list_a = new LinkedList<Character>();
        list_a.add('C');
        list_a.add('D');
        LinkedList<Character> list_b = new LinkedList<Character>();
        list_b.add('D');
        list_b.add('E');
        LinkedList<Character> list_c = new LinkedList<Character>();
        list_c.add('E');
        LinkedList<Character> list_d = new LinkedList<Character>();
        list_c.add('E');

        //构造图
        graph = new HashMap<Character, LinkedList<Character>>();
        graph.put('S', list_s);
        graph.put('A', list_a);
        graph.put('B', list_b);
        graph.put('C', list_c);
        graph.put('D', list_d);
        graph.put('E', new LinkedList<Character>());

        //调用
        dirt = new HashMap<Character, Integer>();
        bfs(graph, dirt, 'S');
    }
}

 3、本题解法：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> result =new ArrayList<>();
        if(root==null) return result;
        Queue<TreeNode> queue=new LinkedList<TreeNode>();
        queue.add(root);
        
        while (!queue.isEmpty()){
            int cur=queue.size();
            double sum=0;
            for(int i=0;i<cur;i++){
                TreeNode treeNode=queue.poll();
                sum+=treeNode.val;
                if(treeNode.left!=null)
                    queue.add(treeNode.left);
                if(treeNode.right!=null)
                    queue.add(treeNode.right);
            }
            result.add(sum/cur);
        }
        return result;
    }
}