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> node_sum;
        vector<int> node_count;
        traverse(root, 1, node_count, node_sum);

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

        return node_sum;
    }

    void traverse(TreeNode *root, int height, vector<int> &node_count, vector<double> &node_sum) {
        if (root == nullptr) {
            return;
        }

        if (height > node_count.size()) {
            node_count.push_back(1);
            node_sum.push_back(root->val);
        } else {
            ++node_count[height - 1];
            node_sum[height - 1] += root->val;
        }

        traverse(root->left, height + 1, node_count, node_sum);
        traverse(root->right, height + 1, node_count, node_sum);
    }
};