[LeetCode March Challange] Day 05 - Average of Levels in Binary Tree

Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. Answers within 10^-5 of the actual answer will be accepted.

Example 1:

Input: root = [3,9,20,null,15,7]
Output: [3.00000,14.50000,11.00000]
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].

Example 2:

Input: root = [3,9,20,15,7]
Output: [3.00000,14.50000,11.00000]

Constraints:

• The number of nodes in the tree is in the range [1, 10^4].
• -2^31 <= Node.val <= 2^(31 - 1)

---

Solution

Time complexity : O(n)
Space complexity : O(n)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root) {
        vector<TreeNode*> q;
        vector<double> ans;
        
        q.push_back(root);
        while(!q.empty()) {
            double avg = .0;
            for (TreeNode* n: q) avg += n->val;
            avg /= q.size();
            ans.push_back(avg);
            
            vector<TreeNode*> next_q;
            for (TreeNode* n: q) {
                if (n->left) next_q.push_back(n->left);
                if (n->right) next_q.push_back(n->right);
            }
            q = next_q;
        }
        
        return ans;
    }
};