Title
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: The range of node’s value is in the range of 32-bit signed integer.
求树每层的平均值,bfs
Solution
//cpp:
/**
* 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) {
queue<TreeNode*> q;
vector<double> v;
int last=1;
if(!root)
return v;
q.push(root);
while(!q.empty()){
double ans=0;
for(int i=0;i<last;i++){
TreeNode* t1=q.front();
ans+=t1->val;
q.pop();
if(t1->left){
q.push(t1->left);
}
if(t1->right){
q.push(t1->right);
}
}
v.push_back(ans/last);
last=q.size();
}
return v;
}
}