LeetCode 637 Average of Levels in Binary Tree（二叉树层序遍历）

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.
题目大意：给出一棵非空的二叉树，求它每一层节点的平均值。

解题思路：非常显然的一道层序遍历。递归版不太好写，没有写出来。

代码如下：

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<double> averageOfLevels(TreeNode* root) {
averageOfLevels(root, ans);
return ans;
}
private:
queue<TreeNode*> que;
vector<double> ans;

void averageOfLevels(TreeNode* root, vector<double>& ans) {
if(root == nullptr) return ;
que.push(root);
TreeNode* tmp;
int cur = 0;
while(cur < que.size()) {
int last = que.size();
double db = 0.0;
while(cur < last) {
tmp = que.front();
db += tmp->val;
que.pop();
if(tmp->left) que.push(tmp->left);
if(tmp->right) que.push(tmp->right);
cur++;
}
cur = 0;
ans.push_back(db / last);
}
}
};