637. Average of Levels in Binary Tree 二叉树每层平均值

Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.
给定一非空二叉树，返回其每层的平均值所构成的数组。
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:
1. The range of node’s value is in the range of 32-bit signed integer.

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思路
【方法1】深度遍历
记录每层的结点总数和结点和。

/** * 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) {        vector<int> count;        vector<double> res;        average(root,0,count,res);        for(int i=0;i<res.size();i++){            res[i]=res[i]/count[i];        }        return res;    }    void average(TreeNode *root, int i, vector<int> &count, vector<double> &sum){        if(!root) return;        if(i<sum.size()){            sum[i]+=root->val;            count[i]++;        }        else{            sum.push_back(1.0*root->val);            count.push_back(1);        }        average(root->left,i+1,count,sum);        average(root->right,i+1,count,sum);    }};

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Double> averageOfLevels(TreeNode root) {        List<Integer> count=new ArrayList<>();        List<Double> res=new ArrayList<>();        average(root,0,count,res);        for(int i=0;i<res.size();i++){            res.set(i,res.get(i)/count.get(i));        }        return res;    }    public void average(TreeNode root, int i, List<Integer> count, List<Double> sum){        if(root==null) return;        if(i<sum.size()){            sum.set(i,sum.get(i)+root.val);            count.set(i,count.get(i)+1);        }        else{            sum.add(1.0*root.val);            count.add(1);        }        average(root.left,i+1,count,sum);        average(root.right,i+1,count,sum);    }}