Eric 的小站

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输入：root = [3,9,20,null,null,15,7]
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输出：[3.00000,14.50000,11.00000]
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解释：第 0 层的平均值为 3,第 1 层的平均值为 14.5,第 2 层的平均值为 11 。
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因此返回 [3, 14.5, 11] 。

1
输入：root = [3,9,20,15,7]
2
输出：[3.00000,14.50000,11.00000]

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/**
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 * Definition for a binary tree node.
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 * struct TreeNode {
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 *     int val;
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 *     TreeNode *left;
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 *     TreeNode *right;
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 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
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 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
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 * right(right) {}
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 * };
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 */
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class Solution {
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public:
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    vector<double> averageOfLevels(TreeNode* root) {
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        vector<TreeNode*> q;
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        vector<double> ans;
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        q.push_back(root);
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        while (!q.empty()) {
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            vector<TreeNode*> vec;
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            double count = 0, sum = 0;
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            for (int i = 0; i < q.size(); i++) {
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                count++;
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                sum += (q[i]->val);
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                TreeNode* node = q[i];
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                if (node->left)
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                    vec.push_back(node->left);
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                if (node->right)
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                    vec.push_back(node->right);
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            }
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            ans.push_back(sum / count);
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            q = move(vec);
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        }
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        return ans;
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    }
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};