637. Average of Levels in Binary Tree

难度: Easy

题目描述

Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. Answers within 10-5 of the actual answer will be accepted.

 

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [3.00000,14.50000,11.00000]
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].

Example 2:

Input: root = [3,9,20,15,7]
Output: [3.00000,14.50000,11.00000]

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

解题思路

代码实现

解决方案

java

/**
 * Definition for a binary tree node.
 * public 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;
 * }
 * }
 */
class Solution {
    public List<Double> averageOfLevels(TreeNode root) {

        if (root == null) {
            return new ArrayList<>();
        }

        LinkedList<TreeNode> cur = new LinkedList<>();
        cur.add(root);
        List<Double> res = new ArrayList<>();
        res.add((double) root.val);
        LinkedList<TreeNode> next = new LinkedList<>();
        while (!cur.isEmpty()) {
            TreeNode first = cur.removeFirst();

            if (first != null) {
                if (first.left != null) {
                    next.add(first.left);
                }
                if (first.right != null) {
                    next.add(first.right);
                }

            }
            if (cur.size() == 0) {

                cur = next;
                if (cur.size() > 0) {
                    res.add(avg(cur));
                }

                next = new LinkedList<>();
            }

        }

        return res;

    }

    private double avg(LinkedList<TreeNode> cur) {
        if (cur.isEmpty()) {
            return 0D;
        }
        double sum = 0D;
        for (TreeNode t : cur) {
            sum += t.val;
        }
        return sum / cur.size();
    }
}