LeetCode-in-Go
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
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func zigzagLevelOrder(root *TreeNode) [][]int {
var result [][]int
if root == nil {
return result
}
queue := []*TreeNode{root, nil}
zig := true
var level []int
for len(queue) > 0 {
node := queue[0]
queue = queue[1:]
for node != nil {
if zig {
level = append(level, node.Val)
} else {
level = append([]int{node.Val}, level...)
}
if node.Left != nil {
queue = append(queue, node.Left)
}
if node.Right != nil {
queue = append(queue, node.Right)
}
if len(queue) > 0 {
node = queue[0]
queue = queue[1:]
} else {
node = nil
}
}
result = append(result, level)
zig = !zig
level = []int{}
if len(queue) > 0 {
queue = append(queue, nil)
}
}
return result
}