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:

  1. The range of node's value is in the range of 32-bit signed integer

    /**
    * 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<double>ret;
    double sum=0,count=0;
    queue<TreeNode*>q;
    q.push(root);
    q.push(nullptr);
    while(!q.empty())
    {
    TreeNode*t=q.front();
    q.pop();
    if(t==nullptr)
    {
    ret.push_back(sum/count);
    sum=0;
    count=0;
    if(!q.empty())
    q.push(nullptr);
    }
    else
    {
    sum+=t->val;
    count++;
    if(t->left)
    q.push(t->left);
    if(t->right)
    q.push(t->right);
    }
    }
    return ret;
    }
    };
    //用nullptr分层,遇到nullptr表示一层结束,算出这一层的平均值添加到数组

posted on 2017-08-01 11:33  王早  阅读(164)  评论(0编辑  收藏  举报