637. Average of Levels in Binary Tree

Solution

/**
 * 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) {}
 * };
 */
#define ll long long

class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root)
    {
        vector<double> res;
        queue<TreeNode*> q{{root}};
        while (!q.empty())
        {            
            int qSize = q.size();
            double sum = 0;
            for (int i = 0; i < qSize; ++i)
            {
                TreeNode* node = q.front(); q.pop();
                sum += node->val;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            res.push_back(sum / qSize);
        }
        return res;
    }
};

• 時間複雜度：
  $O(n)$
• 空間複雜度：
  $O(m)$