给你二叉树的根节点 root ，返回其节点值的 锯齿形层序遍历 。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。
示例 1：
输入：root = [3,9,20,null,null,15,7] 输出：[[3],[20,9],[15,7]]
示例 2：
输入：root = [1] 输出：[[1]]
示例 3：
输入：root = [] 输出：[]
提示：
- 树中节点数目在范围
[0, 2000]内 -100 <= Node.val <= 100
方法一：BFS
为了实现锯齿形层序遍历，需要在层序遍历的基础上增加一个标志位 left，用于标记当前层的节点值的顺序。如果 left 为 true，则当前层的节点值按照从左到右的顺序存入结果数组 ans 中；如果 left 为 false，则当前层的节点值按照从右到左的顺序存入结果数组 ans 中。
时间复杂度
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
ans = []
if root is None:
return ans
q = deque([root])
ans = []
left = 1
while q:
t = []
for _ in range(len(q)):
node = q.popleft()
t.append(node.val)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
ans.append(t if left else t[::-1])
left ^= 1
return ans/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
if (root == null) {
return ans;
}
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
boolean left = true;
while (!q.isEmpty()) {
List<Integer> t = new ArrayList<>();
for (int n = q.size(); n > 0; --n) {
TreeNode node = q.poll();
t.add(node.val);
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
if (!left) {
Collections.reverse(t);
}
ans.add(t);
left = !left;
}
return ans;
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> ans;
if (!root) return ans;
queue<TreeNode*> q{{root}};
int left = 1;
while (!q.empty()) {
vector<int> t;
for (int n = q.size(); n; --n) {
auto node = q.front();
q.pop();
t.emplace_back(node->val);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
if (!left) reverse(t.begin(), t.end());
ans.emplace_back(t);
left ^= 1;
}
return ans;
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func zigzagLevelOrder(root *TreeNode) (ans [][]int) {
if root == nil {
return
}
q := []*TreeNode{root}
left := true
for len(q) > 0 {
t := []int{}
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
t = append(t, node.Val)
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
if !left {
for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
t[i], t[j] = t[j], t[i]
}
}
ans = append(ans, t)
left = !left
}
return
}/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var zigzagLevelOrder = function (root) {
const ans = [];
if (!root) {
return ans;
}
const q = [root];
let left = 1;
while (q.length) {
const t = [];
for (let n = q.length; n; --n) {
const node = q.shift();
t.push(node.val);
if (node.left) {
q.push(node.left);
}
if (node.right) {
q.push(node.right);
}
}
if (!left) {
t.reverse();
}
ans.push(t);
left ^= 1;
}
return ans;
};/**
* 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 res = [];
if (root == null) {
return res;
}
let isDesc = false;
const queue = [root];
while (queue.length !== 0) {
const arr = queue.slice().map(() => {
const { val, left, right } = queue.shift();
left && queue.push(left);
right && queue.push(right);
return val;
});
res.push(isDesc ? arr.reverse() : arr);
isDesc = !isDesc;
}
return res;
}// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
pub fn zigzag_level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
let mut res = vec![];
if root.is_none() {
return res;
}
let mut is_desc = false;
let mut q = VecDeque::new();
q.push_back(root);
while !q.is_empty() {
let mut arr = vec![];
for _ in 0..q.len() {
if let Some(node) = q.pop_front().unwrap() {
let mut node = node.borrow_mut();
arr.push(node.val);
if node.left.is_some() {
q.push_back(node.left.take());
}
if node.right.is_some() {
q.push_back(node.right.take());
}
}
}
if is_desc {
arr.reverse();
}
is_desc = !is_desc;
res.push(arr);
}
res
}
}