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[9/2] 637. Average of Levels in Binary Tree

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작성일 2022.09.02 14:24

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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

Accepted

348,662

Submissions

489,053

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링크

https://leetcode.com/problems/average-of-levels-in-binary-tree/ 1064 회 연결

익명
작성일 2022.09.02 14:24

Runtime: 97 ms, faster than 23.20% of Python3 online submissions for Average of Levels in Binary Tree.
Memory Usage: 16.5 MB, less than 45.98% of Python3 online submissions for Average of Levels in Binary Tree.

class Solution:
    def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:
        # bfs solution
        queue = deque([root])
        result = []
        
        while queue:
            this_level = []
            for i in range(len(queue)):
                node = queue.popleft()
                this_level.append(node.val)
                if node.left: queue.append(node.left)
                if node.right: queue.append(node.right)
            result.append(sum(this_level)/len(this_level))
        return result

익명
작성일 2022.09.05 15:31

/**
 * 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) {
        queue<TreeNode*> q;
        vector<double> ans;
        int current = 0;
        int next = 0;
        double sum = 0;
        int count = 0;
        TreeNode* node = nullptr;
        
        if(root)
        {
            q.push(root);
            current = 1;
            while(!q.empty()) {
                node = q.front();
                q.pop();
                
                if(node->left) {
                    q.push(node->left);
                    next++;
                }
                if(node->right) {
                    q.push(node->right);
                    next++;
                }
                sum += node->val;
                
                count++;
                
                if(count == current) {
                    ans.push_back(sum/count);
                    sum = 0;
                    count = 0;
                    current = next;
                    next = 0;
                }
            }
            
        }
        
        
        
        return ans;
    }
};

Runtime: 38 ms, faster than 15.57% of C++ online submissions for Average of Levels in Binary Tree.
Memory Usage: 22.5 MB, less than 29.85% of C++ online submissions for Average of Levels in Binary Tree.
[code=c++]/**
 * 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) {
        queue<TreeNode*> q;
        vector<double> ans;
        int current = 0;
        int next = 0;
        double sum = 0;
        int count = 0;
        TreeNode* node = nullptr;
        
        if(root)
        {
            q.push(root);
            current = 1;
            while(!q.empty()) {
                node = q.front();
                q.pop();
                
                if(node->left) {
                    q.push(node->left);
                    next++;
                }
                if(node->right) {
                    q.push(node->right);
                    next++;
                }
                sum += node->val;
                
                count++;
                
                if(count == current) {
                    ans.push_back(sum/count);
                    sum = 0;
                    count = 0;
                    current = next;
                    next = 0;
                }
            }
            
        }
        
        
        
        return ans;
    }
};[/code]