637. Average of Levels in Binary Tree

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.

思路:用队列实现按层遍历二叉树.关键是如何确定哪些结点在同一层.用队列解决:从根开始往下,每次往队列中填充一层的node,第一次即根root,然后计算此时队列大小count,这就是第一层的节点数.for循环处理第一层所有结点,将其左右子节点入栈(若存在的话),并累加该节点的val,循环结束后计算第一层的平均值.此时队列中全部为第二层的结点.依此类推,记第二层结点数为n2,从队列中弹出两个队首元素做处理,然后将第三层结点入队尾......直至队列为空.

/**
 * 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) {
        queue<TreeNode*> q;//结点队列
        vector<double> result;
        q.push(root);//从根开始
        while (!q.empty()) {
            int count = q.size();//记录该层大小
            double sum = 0.0;//累加和
            for (int i = 0; i < count; i++) {
                TreeNode* tmp = q.front();//取队首
                q.pop();//弹出队首
                sum += tmp->val;
                if (tmp->left) q.push(tmp->left);//若存在,则左子入队
                if (tmp->right) q.push(tmp->right);//若存在,则右子入队
            }
            result.push_back(sum/count);
        }
        return result;
    }
};