Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

Example 1:

<b>Input:</b>
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.

分析

树的层序遍历

/**
* 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> res;
double sum=0,cnt=0;
queue<TreeNode*> q;
q.push(root);
TreeNode* last=root;
TreeNode* tail;
while(!q.empty()){
TreeNode* t=q.front();
q.pop();
sum+=t->val;
cnt++;
if(t->left!=NULL){
q.push(t->left);
tail=t->left;
}
if(t->right!=NULL){
q.push(t->right);
tail=t->right;
}
if(t==last){
last=tail;
res.push_back(sum/cnt);
sum=cnt=0;
}
}
return res;
}
};

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