Average of Levels in Binary Tree

LeetCode 637 | Difficulty: Easy

Easy

Problem Description

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, 10^4]`.

- `-2^31 <= Node.val <= 2^31 - 1`

Topics: Tree, Depth-First Search, Breadth-First Search, Binary Tree

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Approach

Tree DFS

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left → go right → combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

Tree BFS (Level-Order)

Use a queue to process the tree level by level. At each level, process all nodes in the queue, then add their children. Track the level size to know when one level ends and the next begins.

When to use

Level-order traversal, level-based aggregation, right/left side view.

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Solutions

Solution 1: C# (Best: 328 ms)

Metric	Value
Runtime	328 ms
Memory	N/A
Date	2018-07-13

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public IList<double> AverageOfLevels(TreeNode root) {
        Queue<TreeNode> level = new Queue<TreeNode>();
    List<double> result = new List<double>();
    if (root == null) return result;
    level.Enqueue(root);
    while (level.Count != 0)
    {
        int rowCount = level.Count;
        double sum = 0.0;
        for (int i = 0; i < rowCount; i++)
        {
            var front = level.Dequeue();
            if (front.left != null) level.Enqueue(front.left);
            if (front.right != null) level.Enqueue(front.right);
            sum += front.val;
        }
        result.Add(sum/(double)rowCount);
    }
    return result;
    }
}

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Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.