637. Average of Levels in Binary Tree

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637. Average of Levels in Binary Tree

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.

Example 1:

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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:

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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

Hints/Notes

  • binary tree

Solution

Language: C++

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/**
 * 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<double> res;
        queue<TreeNode*> q;
        if (root)
            q.push(root);
        while (!q.empty()) {
            int n = q.size();
            double sum = 0;
            for (int i = 0; i < n; i++) {
                TreeNode* cur = q.front();
                q.pop();
                sum += cur->val;
                if (cur->left)
                    q.push(cur->left);
                if (cur->right)
                    q.push(cur->right);
            }
            res.push_back(sum / n);
        }
        return res;
    }
};