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.
/**
* 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> res;
void bfs(TreeNode* root){
deque<TreeNode*> d(NULL);
d.push_back(root);
double tmp = 0;
int fatherNum = 1, childNum = 0, cnt = 0;
while(1){
if(root->left != NULL){
d.push_back(root->left);
childNum++;
}
if(root->right != NULL){
d.push_back(root->right);
childNum++;
}
tmp += d[0]->val;
d.pop_front();
cnt++;
if(cnt == fatherNum){
res.push_back(tmp / cnt);
tmp = 0;
cnt = 0;
fatherNum = childNum;
childNum = 0;
}
if(d.empty())
break;
root = d[0];
}
}
vector<double> averageOfLevels(TreeNode* root) {
if(root == NULL) return NULL;
bfs(root);
return res;
}
};
使用广度搜索,然后记录父节点个数和儿子节点个数即可