637. 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,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 [$1,10^4$].
• $-2^{31}<=Node.val<=2^{31}-1$

From: LeetCode
Link: 637. Average of Levels in Binary Tree

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Solution:

Ideas：

1. Initialize a queue to store nodes at the current level.

2. Add the root node to the queue.

3. While there are nodes in the queue:
   a. Record the number of nodes at the current level (i.e., the size of the queue).

   b. Initialize two variables: sum for the sum of node values at the current level, and count for the number of nodes.

   c. For each node at the current level:

   • i. Dequeue the node.
   • ii. Add its value to sum.
   • iii. Enqueue its left child and right child (if they exist).
   • iv. Increment count.

   d. Compute the average for the current level by dividing sum by count. Add this average to the result list.

4. Return the result list.

Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
double* averageOfLevels(struct TreeNode* root, int* returnSize){
    // Edge case: if the tree is empty
    if (!root) {
        *returnSize = 0;
        return NULL;
    }
    
    struct TreeNode** queue = (struct TreeNode**)malloc(10000 * sizeof(struct TreeNode*));
    double* result = (double*)malloc(10000 * sizeof(double));
    int front = 0, rear = 0, levelCount, resultIndex = 0;
    double sum;
    
    queue[rear++] = root; // enqueue root
    
    while (front < rear) {
        levelCount = rear - front; // nodes at current level
        sum = 0;
        for (int i = 0; i < levelCount; i++) {
            struct TreeNode* currentNode = queue[front++];
            sum += currentNode->val;
            if (currentNode->left) queue[rear++] = currentNode->left;
            if (currentNode->right) queue[rear++] = currentNode->right;
        }
        result[resultIndex++] = sum / levelCount;
    }
    
    *returnSize = resultIndex;
    free(queue);
    return result;
}