Average of Levels in Binary Tree

• LeetCode: Average of Levels in Binary Tree
• Problem:

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

• Note:

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

• Examples:

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].

• Analysis:

For every level, we need sum of all nodes and the number of all noded.

We can use "Level Order Traversal" to get this target.

• Code - 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> results = new ArrayList<Double>();
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        if (root == null) {
            results.add(0.0);
            return results;
        }
        queue.add(root);
        while (!queue.isEmpty()) {
            int len = queue.size();
            double avg = 0.0;
            for (int i = 0; i < len; i++) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
                avg += node.val;
            }
            avg /= (double)len;
            results.add(avg);
        }
        return results;
    }
}

• Code - C++:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<double> averageOfLevels(TreeNode* root) {
        vector<double> result;
        queue<TreeNode*> queue;
        if (root == NULL) {
            return result;
        }
        queue.push(root);
        while (!queue.empty()) {
            int size = queue.size();
            double val = 0.0;
            for (int i = 0; i < size; i++) {
                TreeNode* node = queue.front();
                val += node->val;
                if (node->left != NULL) {
                    queue.push(node->left);
                }
                if (node->right != NULL) {
                    queue.push(node->right);
                }
                queue.pop();
            }
            result.push_back(val / (double) size);
        }
        return result;
    }
};

• Code - Python:

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

import Queue

class Solution(object):
    def averageOfLevels(self, root):
        """
        :type root: TreeNode
        :rtype: List[float]
        """
        result = []
        q = Queue.Queue()
        if root == None:
            return result
        q.put(root)
        while q.qsize() != 0:
            size = q.qsize()
            val = 0.0
            for i in range(size):
                node = q.get()
                val += node.val
                if node.left != None:
                    q.put(node.left)
                if node.right != None:
                    q.put(node.right)
            result.append(val / (0.0 + size))
        return result

• Time Complexity: O(N), N is the number of nodes in the tree.
• Space Complexity: O(LogN), LogN is the total level of the tree.