LeetCode637. Average of Levels in Binary Tree

文章目录

    • 一、题目
    • 二、题解

一、题目

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, 104].
-231 <= Node.val <= 231 - 1

二、题解

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

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