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:

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

给定一颗二叉树，求出二叉树每一层节点的平均值，第一反应是使用二叉树的层次遍历，先回忆一下二叉树的层次遍历

void level_tree(bintree t){  
    Queue q;  
    bintree temp;  
    if(!t){  
        printf("the tree is empty
");  
        return ;  
    }  
    q.add(t);  
    while(!q.isEmpty){  
        t=poll(q);   //出队
        printf("%c ",t.val);  
        if(t.left != null){  
             q.add(t.left);
        }  
        if(t.right != null){  
            q.add(t.right); 
        }  
    }  
} 

 

有了模型之后，并不能直接使用，因为这一题要的是每一层的均值，我们需要记录下某一层的节点都有哪些，层次遍历时候，当开始遍历某一层时，队列的大小就是该层的节点数。

public List<Double> averageOfLevels(TreeNode root) {
            List < Double > res = new ArrayList < > ();
            Queue<TreeNode> q=new LinkedList<>();
            q.add(root);
            while(!q.isEmpty())
            {
                int n = q.size();
                double sum = 0.0;
                for(int i = 0; i<n;i++) //这个地方设计的比较巧妙，用来计算一层的节点
                {
                    TreeNode x = q.poll();
                    sum += x.val;
                    if(x.left != null)
                        q.add(x.left);
                    if(x.right != null)
                        q.add(x.right);//同q.offer
                    
                }
                res.add(sum / n);
            }
            return res; 
    }