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:
3
/ \
9 20
/ \
15 7
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
BFS level by level
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# 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]]:
if not root:
return []
res = []
direction = 1
level = [root]
while level:
res.append([n.val for n in level][::direction])
direction *= -1
level = [
child
for node in level
for child in (node.left, node.right)
if child
]
return res
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/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func traverseTree(root *TreeNode, level int, result *[][]int) {
if root == nil {
return
}
if level > len(*result)-1 {
*result = append(*result, []int{})
}
(*result)[level] = append((*result)[level], root.Val)
traverseTree(root.Left, level+1, result)
traverseTree(root.Right, level+1, result)
}
func reverse(nums []int) {
for i := 0; i < len(nums)/2; i++ {
j := len(nums) - i - 1
nums[i], nums[j] = nums[j], nums[i]
}
}
func zigzagLevelOrder(root *TreeNode) [][]int {
var result [][]int
traverseTree(root, 0, &result)
for i := 1; i < len(result); i += 2 {
reverse(result[i])
}
return result
}
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