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题目

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.

讲解

这题跟429. N-ary Tree Level Order Traversal很像。都是广度优先遍历树(也就是按层级遍历树)。

java代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<Double> result = new ArrayList<>();
    public List<Double> averageOfLevels(TreeNode root) {
        if(root==null){
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int children_num=1;
        while(!queue.isEmpty()){
            double sum = 0;
            int count=0;
            for(int i=0;i<children_num;i++){
                TreeNode now = queue.poll();
                sum += now.val;
                if(now.left!=null){
                    count++;
                    queue.offer(now.left);
                }
                if(now.right!=null){
                    count++;
                    queue.offer(now.right);
                }
            }
            sum /= children_num;
            result.add(sum);
            children_num = count;
        }
        return result;
    }
}

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