Leetcode 637. Average of Levels in Binary Tree

标签：nat   for   else   tor   form   eve   ==   ble   pre   

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

Leetcode 637. Average of Levels in Binary Tree

标签：nat   for   else   tor   form   eve   ==   ble   pre   

原文地址：http://www.cnblogs.com/wangzao2333/p/7267588.html