给定一个二叉树，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）。
例如：
给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回锯齿形层序遍历如下：
[ [3], [20,9], [15,7] ]
# 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: TreeNode) -> List[List[int]]:
if root is None:
return []
left, ans = False, []
q = deque([root])
while q:
n = len(q)
t = []
for _ in range(n):
node = q.popleft()
t.append(node.val)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
if left:
t.reverse()
ans.append(t)
left = not left
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) {
if (root == null) {
return Collections.emptyList();
}
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
boolean left = false;
List<List<Integer>> ans = new ArrayList<>();
while (!q.isEmpty()) {
List<Integer> t = new ArrayList<>();
for (int i = 0, n = q.size(); i < n; ++i) {
TreeNode node = q.pollFirst();
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) {
if (!root) return {};
queue<TreeNode*> q{{root}};
vector<vector<int>> ans;
bool left = false;
while (!q.empty())
{
vector<int> t;
for (int i = 0, n = q.size(); i < n; ++i)
{
auto node = q.front();
q.pop();
t.push_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.push_back(t);
left = !left;
}
return ans;
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func zigzagLevelOrder(root *TreeNode) [][]int {
if root == nil {
return nil
}
var ans [][]int
var q = []*TreeNode{root}
left := false
for len(q) > 0 {
var t []int
n := len(q)
for i := 0; i < n; i++ {
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 {
i, j := 0, n-1
for i < j {
t[i], t[j] = t[j], t[i]
i++
j--
}
}
ans = append(ans, t)
left = !left
}
return ans
}/**
* 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) {
if (!root) {
return [];
}
let ans = [];
let q = [root];
let left = false;
while (q.length) {
let t = [];
for (let i = 0, n = q.length; i < n; ++i) {
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 = !left;
}
return ans;
};