637. Average of Levels in Binary Tree

最新推荐文章于 2024-08-26 10:59:05 发布

转载最新推荐文章于 2024-08-26 10:59:05 发布 · 61 阅读

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原文链接：http://www.cnblogs.com/gsz-/p/9520345.html

本文介绍了一种算法，用于求解给定非空二叉树各层节点值的平均数，并通过队列实现逐层遍历。示例中输入为特定二叉树结构，输出则是各层节点值的平均数数组。

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题目描述：

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:

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

解题思路：

使用queue，遍历树的每一层，确保每次queue中只保存了一层的节点。

代码：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<double> averageOfLevels(TreeNode* root) {
13         vector<double> ret;
14         queue<TreeNode*> nodes;
15         nodes.push(root);
16         while (!nodes.empty()) {
17             double sum = 0;
18             int size = nodes.size();
19             for (int i = 0; i < size; ++i) {
20                 TreeNode* tmp = nodes.front();
21                 nodes.pop();
22                 sum += tmp->val;
23                 if (tmp->left)
24                     nodes.push(tmp->left);
25                 if (tmp->right)
26                     nodes.push(tmp->right);
27             }
28             ret.push_back(sum / size);
29         }
30         return ret;
31     }
32 };