Published Feb 24, 2022
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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).
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]
Input: root = [1]
Output: [[1]]
Input: root = []
Output: []
- The number of nodes in the tree is in the range [0, 2000].
- -100 <= Node.val <= 100
/**
* 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[][] {
let res: number[][] = [];
let isReverse: boolean = false;
let queue: TreeNode[] = [];
queue.push(root);
while(queue.length !== 0) {
let temp: number[] = [];
const LEN: number = queue.length;
for(let i = 0; i < LEN; i++){
const front: TreeNode = queue.shift();
if(front !== null){
temp.push(front.val);
queue.push(front.left);
queue.push(front.right);
}
}
if(temp.length !== 0) {
if(isReverse){
temp.reverse()
}
isReverse = !isReverse;
res.push(temp)
}
}
return res;
};