[LeetCode] 637. Average of Levels in Binary Tree 二叉树的层平均值

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.

给一个非空二叉树，返回每层的平均值组成的数组。

解法：BFS迭代，层序遍历，计算每层的平均值记录到res数组，最后返回res数组。

Java:

public List<Double> averageOfLevels(TreeNode root) {
    List<Double> result = new ArrayList<>();
    Queue<TreeNode> q = new LinkedList<>();
    
    if(root == null) return result;
    q.add(root);
    while(!q.isEmpty()) {
        int n = q.size();
        double sum = 0.0;
        for(int i = 0; i < n; i++) {
            TreeNode node = q.poll();
            sum += node.val;
            if(node.left != null) q.offer(node.left);
            if(node.right != null) q.offer(node.right);
        }
        result.add(sum / n);
    }
    return result;
}　　

Java: BFS

public List<Double> averageOfLevels(TreeNode root) {
        List<Double> list = new LinkedList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int count = queue.size();
            double sum = 0;
            for (int i = 0; i < count; i++) {
                TreeNode cur = queue.poll();
                sum += cur.val;
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
            }
            list.add(sum / count);
        }
        return list;
    }

Java: DFS　　

class Node {
        double sum;
        int count;
        Node (double d, int c) {
            sum = d;
            count = c;
        }
    }
    public List<Double> averageOfLevels(TreeNode root) {
        List<Node> temp = new ArrayList<>();
        helper(root, temp, 0);
        List<Double> result = new LinkedList<>();
        for (int i = 0; i < temp.size(); i++) {
            result.add(temp.get(i).sum / temp.get(i).count);
        }
        return result;
    }
    public void helper(TreeNode root, List<Node> temp, int level) {
        if (root == null) return;
        if (level == temp.size()) {
            Node node = new Node((double)root.val, 1);
            temp.add(node);
        } else {
            temp.get(level).sum += root.val;
            temp.get(level).count++;
        }
        helper(root.left, temp, level + 1);
        helper(root.right, temp, level + 1);
    }　　

Python:

# Time:  O(n)
# Space: O(h)
class Solution(object):
    def averageOfLevels(self, root):
        """
        :type root: TreeNode
        :rtype: List[float]
        """
        result = []
        q = [root]
        while q:
            total, count = 0, 0
            next_q = []
            for n in q:
                total += n.val
                count += 1
                if n.left:
                    next_q.append(n.left)
                if n.right:
                    next_q.append(n.right)
            q = next_q
            result.append(float(total) / count)
        return result

Python:

class Solution(object):
    def averageOfLevels(self, root):
        """
        :type root: TreeNode
        :rtype: List[float]
        """
        if root is None:
            return []
        result, current = [], [root]
        while current:
            next_level, vals,  = [], 0.0
            counts = len(current)
            for node in current:
                vals += node.val
                if node.left:
                    next_level.append(node.left)
                if node.right:
                    next_level.append(node.right)                    
            current = next_level
            result.append(vals / counts)
            
        return result 　　

Python: wo　　

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def averageOfLevels(self, root):
        """
        :type root: TreeNode
        :rtype: List[float]
        """
                
        res = []
        self.helper([root], res)
        return res
                
    def helper(self, q, res):
        sm = 0.0
        counts = 0
        next_level = []
        while q:
            node = q.pop()
            counts += 1
            sm += node.val
            if node.left:
                next_level.append(node.left)
            if node.right:
                next_level.append(node.right)
        res.append(sm / counts)
        if next_level:
            self.helper(next_level, res)

C++:

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

C++:

class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root) {
        vector<double> res, cnt;
        helper(root, 0, cnt, res);
        for (int i = 0; i < res.size(); ++i) {
            res[i] /= cnt[i];
        }
        return res;
    }
    void helper(TreeNode* node, int level, vector<double>& cnt, vector<double>& res) {
        if (!node) return;
        if (res.size() <= level) {
            res.push_back(0);
            cnt.push_back(0);
        }
        res[level] += node->val;
        ++cnt[level];
        helper(node->left, level + 1, cnt, res);
        helper(node->right, level + 1, cnt, res);
    }
};

　　

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[LeetCode] 102. Binary Tree Level Order Traversal 二叉树层序遍历

[LeetCode] 107. Binary Tree Level Order Traversal II 二叉树层序遍历 II

[LeetCode] 199. Binary Tree Right Side View 二叉树的右侧视图

　　

　　

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