639. 解码方法II
- 题号:639
- 来源:LeetCode
- 难度:简单
- 标签:
BFS树队列 - 语言:TypeScript
- 解法:1 个
- 作者:lmliheng
- 最近更新:2026-09-27
TypeScript · O(n) 时间 · O(n) 空间 · 更新于 2026-09-27
层序遍历求每层均值,与文件名不符
typescript
/**
* @difficulty easy
* @tags BFS,树,队列
* @time O(n)
* @space O(n)
* @note 层序遍历求每层均值,与文件名不符
* @639. 解码方法II
*/
function TreeNode(val: number, left?: TreeNode | null, right?: TreeNode | null) {
this.val = (val === undefined ? 0 : val)
this.left = (left === undefined ? null : left)
this.right = (right === undefined ? null : right)
}
let root = new TreeNode(3, new TreeNode(9, new TreeNode(1)), new TreeNode(20, new TreeNode(15), new TreeNode(7)))
let res: number[] = []
const bfs = (root: TreeNode) => {
let quene: TreeNode[] = [root]
while (quene.length) {
let sum = 0
for (let i = 0; i < quene.length; i++) {
sum += quene[i].val
}
res.push(sum / quene.length)
// 每一轮遍历时,将队列中的节点全部取出,
// 计算这些节点的数量以及它们的节点值之和,并计算这些节点的平均值,
// 然后将这些节点的全部非空子节点加入队列,重复上述操作直到队列为空,遍历结束。
let queneLength = quene.length
for (let i = 0; i < queneLength; i++) {
let node = quene.shift()!
if (node.left) {
quene.push(node.left)
}
if (node.right) {
quene.push(node.right)
}
}
}
}
bfs(root)
console.log(res)在 GitHub 上查看题目所在目录:lmliheng/algorithm