637. 二叉树的层平均值

给定一个非空二叉树, 返回一个由每层节点平均值组成的数组。

示例 1：

输入：
3
/
9 20
/
15 7
输出：[3, 14.5, 11]
解释：
第 0 层的平均值是 3 , 第1层是 14.5 , 第2层是 11 。因此返回 [3, 14.5, 11] 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/average-of-levels-in-binary-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

广度优先搜索

import java.util.*;

class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        if (root == null) {
            return Collections.emptyList();
        }
        List<Double> ret = new ArrayList<>();

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            double sum = 0;
            int size = queue.size();
            for (int i = 1; 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);
                }
            }
            ret.add(sum / size);
        }
        return ret;
    }
}


class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode() {
    }

    TreeNode(int val) {
        this.val = val;
    }

    TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

深度优先搜索

import java.util.*;

class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Integer> counts = new ArrayList<>();
        List<Double> sums = new ArrayList<>();
        dfs(root, 0, counts, sums);
        List<Double> averages = new ArrayList<>();
        int size = sums.size();
        for (int i = 0; i < size; i++) {
            averages.add(sums.get(i) / counts.get(i));
        }
        return averages;
    }

    public void dfs(TreeNode root, int level, List<Integer> counts, List<Double> sums) {
        if (root == null) {
            return;
        }
        if (level < sums.size()) {
            sums.set(level, sums.get(level) + root.val);
            counts.set(level, counts.get(level) + 1);
        } else {
            sums.add(1.0 * root.val);
            counts.add(1);
        }
        dfs(root.left, level + 1, counts, sums);
        dfs(root.right, level + 1, counts, sums);
    }
}


class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode() {
    }

    TreeNode(int val) {
        this.val = val;
    }

    TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}