1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63

// @Title: 从上到下打印二叉树 III (从上到下打印二叉树 III LCOF)
// @Author: 15816537946@163.com
// @Date: 2022-02-11 23:15:00
// @Runtime: 0 ms
// @Memory: 2.7 MB
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrder(root *TreeNode) [][]int {
    if root == nil {
        return nil
    }

    res := [][]int{}
    queue := []*TreeNode{root}

    for len(queue) != 0 {
        size := len(queue)
        curLevel := make([]int,size)

        if len(res)%2 == 1{
        for i:=0;i<size;i++ {
            node := queue[0]
            queue = queue[1:]
            
                if node != nil {
                    curLevel[size-i-1] = node.Val
                }
                if node.Left != nil {
                    queue = append(queue, node.Left)
                }
                if node.Right != nil {
                    queue = append(queue, node.Right)
                }
            }
            res = append(res,curLevel)
        } else {
            for i :=0;i<size;i++ {
                node := queue[0]
                queue = queue[1:]

                if node != nil {
                    curLevel[i] = node.Val
                }
                if node.Left != nil {
                    queue = append(queue, node.Left)
                }
                if node.Right != nil {
                    queue = append(queue, node.Right)
                }
            }

            res = append(res, curLevel)
        }
    }

    return res
}