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Tree BFS

Guide 5 of 6 · Path 5 of 6

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Demo preview: These solutions pass the automated test suite but have not been human-reviewed. Treat trade-offs and prose as draft.

Easytree-bfs

Average of Levels in Binary Tree

Problem (restated)

Return the average value of the nodes on each level as an array of doubles.

Intuition

Level BFS; sum/count for each level.

Approaches

Level BFS average

Tested only
Time O(n)Space O(w)

Idea. For each level size, accumulate sum, push sum/size.

Walkthrough. [3,9,20,null,null,15,7] → [3, 14.5, 11].

Trade-offs. Same skeleton as level order; only aggregate differs.

Solution
export class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val = 0, left: TreeNode | null = null, right: TreeNode | null = null) {
    this.val = val; this.left = left; this.right = right;
  }
}

export function averageOfLevels(root: TreeNode | null): number[] {
  if (!root) return [];
  const res: number[] = [];
  const q: TreeNode[] = [root];
  while (q.length) {
    const size = q.length;
    let sum = 0;
    for (let i = 0; i < size; i++) {
      const n = q.shift()!;
      sum += n.val;
      if (n.left) q.push(n.left);
      if (n.right) q.push(n.right);
    }
    res.push(sum / size);
  }
  return res;
}

export class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val = 0, left: TreeNode | null = null, right: TreeNode | null = null) {
    this.val = val; this.left = left; this.right = right;
  }
}

export function averageOfLevels(root: TreeNode | null): number[] {
  if (!root) return [];
  const res: number[] = [];
  const q: TreeNode[] = [root];
  while (q.length) {
    const size = q.length;
    let sum = 0;
    for (let i = 0; i < size; i++) {
      const n = q.shift()!;
      sum += n.val;
      if (n.left) q.push(n.left);
      if (n.right) q.push(n.right);
    }
    res.push(sum / size);
  }
  return res;
}

Template connection

Tree BFS per-level aggregate.

Reflection