LeetCode 637. Average of Levels in Binary Tree

原题链接在这里：https://leetcode.com/problems/average-of-levels-in-binary-tree/discuss/

题目：

Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

Example 1:

Input:
    3
   / \
  9  20
    /  \
   15   7
Output: [3, 14.5, 11]
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].

Note:

1. The range of node's value is in the range of 32-bit signed integer.

题解：

可以采用BFS通过queue的size结算每层level的个数. 挨个poll出来计算sum.

Time Complexity: O(n), n 是node 个数. 

Space: O(n).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<Double>();
        if(root == null){
            return res;
        }
        
        LinkedList<TreeNode> que = new LinkedList<TreeNode>();
        que.add(root);
        while(!que.isEmpty()){
            int count = que.size();
            double sum = 0.0;
            for(int i = 0; i<count; i++){
                TreeNode tn = que.poll();
                sum += tn.val;
                if(tn.left != null){
                    que.add(tn.left);
                }
                if(tn.right != null){
                    que.add(tn.right);
                }
            }
            res.add(sum/count);
        }
        return res;
    }
}

AC 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) {
        vector<double> res;
        if(!root){
            return res;
        }

        queue<TreeNode*> que;
        que.push(root);
        while(!que.empty()){
            int size = que.size();
            double sum = 0.0;
            for(int i = 0; i < size; i++){
                TreeNode* cur = que.front();
                que.pop();
                sum += cur->val;

                if(cur->left){
                    que.push(cur->left);
                }

                if(cur->right){
                    que.push(cur->right);
                }
            }

            res.push_back(sum / size);
        }

        return res;
    }
};

AC Python:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:
        res = []
        if not root:
            return res
        
        que = deque([root])
        while que:
            size = len(que)
            sum = 0
            for i in range(size):
                cur = que.popleft()
                sum += cur.val
                if cur.left:
                    que.append(cur.left)
                if cur.right:
                    que.append(cur.right)
                
            res.append(sum / size)

        return res
        

也可以使用DFS. create 新的class Node 来维护每层的sum 和count.

Time Complexity: O(n). Space: O(logn).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    class Node{
        double sum;
        int count;
        Node(double sum, int count){
            this.sum = sum;
            this.count = count;
        }
    }
    
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> res = new ArrayList<Double>();
        List<Node> nodes = new ArrayList<Node>();
        
        avaerageOfLevelsDfs(root, nodes, 0);
        for(int i = 0; i<nodes.size(); i++){
            res.add(nodes.get(i).sum/nodes.get(i).count);
        }
        return res;
    }
    
    private void avaerageOfLevelsDfs(TreeNode root,  List<Node> nodes, int level){
        if(root == null){
            return;
        }
        
        if(level == nodes.size()){
            nodes.add(new Node(1.0 * root.val, 1));
        }else{
            nodes.get(level).sum += root.val;
            nodes.get(level).count++;
        }
        
        avaerageOfLevelsDfs(root.left, nodes, level+1);
        avaerageOfLevelsDfs(root.right, nodes, level+1);
    }
}