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:
- The range of node's value is in the range of 32-bit signed integer.
二叉树的层平均值。
给定一个非空二叉树, 返回一个由每层节点平均值组成的数组。
题意很简单，思路也类似102题，用BFS做传统的level orver traversal。直接给代码了。
时间O(n)
空间O(n)
Java实现
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode() {} 8 * TreeNode(int val) { this.val = val; } 9 * TreeNode(int val, TreeNode left, TreeNode right) { 10 * this.val = val; 11 * this.left = left; 12 * this.right = right; 13 * } 14 * } 15 */ 16 class Solution { 17 public List<Double> averageOfLevels(TreeNode root) { 18 List<Double> res = new ArrayList<>(); 19 // corner case 20 if (root == null) { 21 return res; 22 } 23 24 // normal case 25 Queue<TreeNode> queue = new LinkedList<>(); 26 queue.offer(root); 27 while (!queue.isEmpty()) { 28 int size = queue.size(); 29 double sum = 0.0; 30 for (int i = 0; i < size; i++) { 31 TreeNode cur = queue.poll(); 32 sum += cur.val; 33 if (cur.left != null) { 34 queue.offer(cur.left); 35 } 36 if (cur.right != null) { 37 queue.offer(cur.right); 38 } 39 } 40 res.add(sum / size); 41 } 42 return res; 43 } 44 }
JavaScript实现
1 /** 2 * @param {TreeNode} root 3 * @return {number[]} 4 */ 5 var averageOfLevels = function (root) { 6 let res = []; 7 // corner case 8 if (root == null) { 9 return res; 10 } 11 12 // normal case 13 let queue = []; 14 queue.push(root); 15 while (queue.length > 0) { 16 let size = queue.length; 17 let sum = 0.0; 18 for (let i = 0; i < size; i++) { 19 let cur = queue.shift(); 20 sum += cur.val; 21 if (cur.left != null) { 22 queue.push(cur.left); 23 } 24 if (cur.right != null) { 25 queue.push(cur.right); 26 } 27 } 28 res.push(sum / size); 29 } 30 return res; 31 };