leetcode 637. Average of Levels in Binary Tree

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   7Output: [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.

简单题，用BFS，唯一注意是sum需要是long类型，不然会溢出。

package leetcode;import java.util.ArrayList;import java.util.LinkedList;import java.util.List;import java.util.Queue;public class Average_of_Levels_in_Binary_Tree_637 {public List<Double> averageOfLevels(TreeNode root) {List<Double> list=new ArrayList<Double>();if(root==null){return list;}Queue<TreeNode> queue=new LinkedList<TreeNode>();queue.offer(root);while(!queue.isEmpty()){int size=queue.size();long sum=0;for(int i=0;i<size;i++){TreeNode node=queue.poll();sum+=node.val;if(node.left!=null){queue.offer(node.left);}if(node.right!=null){queue.offer(node.right);}}double avg=(double)sum/size;list.add(avg);}return list;}public static void main(String[] args) {// TODO Auto-generated method stubAverage_of_Levels_in_Binary_Tree_637 a=new Average_of_Levels_in_Binary_Tree_637();TreeNode root=new TreeNode(2147483647);root.left=new TreeNode(2147483647);root.right=new TreeNode(2147483647);List<Double> list=a.averageOfLevels(root);for(Double d:list){System.out.print(d+" ");}}}

大神DFS解法见：https://leetcode.com/problems/average-of-levels-in-binary-tree/#/solution

其中 i 代表层数。sum这个list中，sum.get(0)代表0层数的和。count这个list中，count.get(0)代表0层数的node个数。

public class Solution {    public List < Double > averageOfLevels(TreeNode root) {        List < Integer > count = new ArrayList < > ();        List < Double > res = new ArrayList < > ();        average(root, 0, res, count);        for (int i = 0; i < res.size(); i++)            res.set(i, res.get(i) / count.get(i));        return res;    }    public void average(TreeNode t, int i, List < Double > sum, List < Integer > count) {        if (t == null)            return;        if (i < sum.size()) {            sum.set(i, sum.get(i) + t.val);            count.set(i, count.get(i) + 1);        } else {            sum.add(1.0 * t.val);            count.add(1);        }        average(t.left, i + 1, sum, count);        average(t.right, i + 1, sum, count);    }}