637. Average of Levels in Binary Tree

2020, Nov 22 One min read

1. Description

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.

2. Example

Example 1

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

Example 2
Input: root = [3,9,20,15,7]
Output: [3.00000,14.50000,11.00000]

3. Constraints

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

4. Solutions

Depth-First Search

n is the number of nodes in root
Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    vector<double> averageOfLevels(TreeNode *root) {
        vector<double> level_node_sums;
        vector<int> level_node_count;
        traverse(root, 0, level_node_sums, level_node_count);

        for (int i = 0, n = level_node_sums.size(); i < n; ++i) {
            level_node_sums[i] /= level_node_count[i];
        }

        return level_node_sums;
    }

private:
    void traverse(
        TreeNode *root,
        int level,
        vector<double> &level_node_sums,
        vector<int> &level_node_count) {
        if (root != nullptr) {
            if (level < level_node_sums.size()) {
                level_node_sums[level] += root->val;
                level_node_count[level]++;
            } else {
                level_node_sums.push_back(root->val);
                level_node_count.push_back(1);
            }

            traverse(root->left, level + 1, level_node_sums, level_node_count);
            traverse(root->right, level + 1, level_node_sums, level_node_count);
        }
    }
};