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.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<double> averageOfLevels(TreeNode* root) { 13 vector<double> vet; 14 if (root == NULL) 15 return vet; 16 stack<TreeNode*> stacknode; 17 stacknode.push(root); 18 while (stacknode.size()) 19 { 20 double sum = 0; 21 int count = stacknode.size(); 22 stack<TreeNode*> stacknode1 = stacknode; 23 while (stacknode.size()) 24 { 25 TreeNode* pNode = stacknode.top(); 26 sum += pNode->val; 27 stacknode.pop(); 28 } 29 vet.push_back(sum / count); 30 while (stacknode1.size()) 31 { 32 TreeNode* pNode1 = stacknode1.top(); 33 if (pNode1->right) 34 stacknode.push(pNode1->right); 35 if (pNode1->left) 36 stacknode.push(pNode1->left); 37 stacknode1.pop(); 38 } 39 40 } 41 return vet; 42 } 43 };
