LeetCode in TypeScript
103. Binary Tree Zigzag Level Order Traversal
Medium
Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]
Example 2:
Input: root = [1]
Output: [[1]]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -100 <= Node.val <= 100
Solution
import { TreeNode } from '../../com_github_leetcode/treenode'
/**
* Definition for a binary tree node.
* 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 zigzagLevelOrder(root: TreeNode | null): number[][] {
const result: number[][] = []
if (root === null) {
return result
}
const q: (TreeNode | null)[] = [root, null]
let zig = true
let level: number[] = []
while (q.length > 0) {
const node = q.shift()
if (node === null) {
result.push(level)
zig = !zig
level = []
if (q.length > 0) {
q.push(null)
}
} else {
if (zig) {
level.push(node.val)
} else {
level.unshift(node.val)
}
if (node.left !== null) {
q.push(node.left)
}
if (node.right !== null) {
q.push(node.right)
}
}
}
return result
}
export { zigzagLevelOrder }