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index.ts
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index.ts
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/**
@see https://leetcode-cn.com/problems/binary-tree-level-order-traversal/
给你一个二叉树,请你返回其按 层序遍历 得到的节点值。 (即逐层地,从左到右访问所有节点)。
示例:
二叉树:[3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回其层序遍历结果:
[
[3],
[9,20],
[15,7]
]
*/
namespace levelOrder{
class TreeNode {
val: number
left: TreeNode | null
right: TreeNode | null
constructor(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)
}
}
function levelOrder(root: TreeNode | null): number[][] {
const result: number[][] = []
const queue: Array<TreeNode | null> = []
//创建一个标识符记录每一层的需要遍历的节点数量
let nums = 1
//创建一个数组来存储每一层的节点值
let tempArr = []
queue.push(root)
while (queue.length > 0) {
const current = queue.shift()
const left = current?.left
const right = current?.right
left && queue.push(left)
right && queue.push(right)
current?.val !== undefined && tempArr.push(current.val)
nums-- //添加一次,减少一个
if (nums === 0) {
nums = queue.length //得到下一层的节点数量
tempArr.length > 0 && result.push(tempArr)
tempArr = [] //重置
}
}
return result
};
}