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   7Output: [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.

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);        }    }};