637. Average of Levels in Binary Tree
方法1：易错点
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.
方法1：
思路：
level order traversal。数据点test到了INT_MAX，IN_MIN注意防止overflow。
易错点
需要用double或者long来防止overflow
class Solution {
public:
vector
<double> averageOfLevels(TreeNode
* root
) {
if (!root
) return {};
vector
<double> result
;
queue
<TreeNode
*> q
;
q
.push(root
);
while (!q
.empty()) {
int sz
= q
.size();
double sum
= 0;
for (int i
= 0; i
< sz
; i
++) {
auto top
= q
.front();
q
.pop();
sum
+= top
-> val
;
if (top
-> left
) q
.push(top
-> left
);
if (top
-> right
) q
.push(top
-> right
);
}
result
.push_back(sum
/ (double) sz
);
}
return result
;
}
};
