剑指 Offer 32 - III. 从上到下打印二叉树 III
mid
请实现一个函数按照之字形顺序打印二叉树,即第一行按照从左到右的顺序打印,第二层按照从右到左的顺序打印,第三行再按照从左到右的顺序打印,其他行以此类推。
例如:
给定二叉树: [3,9,20,null,null,15,7]
,
1
2
3
4
5
| 3
/ \
9 20
/ \
15 7
|
返回其层次遍历结果:
1
2
3
4
5
| [
[3],
[20,9],
[15,7]
]
|
解
隔一个倒序一下
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
| func levelOrder(root *TreeNode) [][]int {
if root == nil {
return [][]int{}
}
result := make([][]int, 0)
queue := make([]*TreeNode, 0)
queue = append(queue, root)
curCount := 1
curLevel := 0
nextCount := 0
isAscending := true
for len(queue) != 0 {
curNodes := make([]int, 0)
for curCount > 0 {
curNode := queue[0]
queue = queue[1:]
curNodes = append(curNodes, curNode.Val)
if curNode.Left != nil {
queue = append(queue, curNode.Left)
nextCount++
}
if curNode.Right != nil {
queue = append(queue, curNode.Right)
nextCount++
}
curCount--
}
if !isAscending {
// 倒序
reversedNodes := make([]int, 0)
for i := len(curNodes) - 1; i >= 0; i-- {
reversedNodes = append(reversedNodes, curNodes[i])
}
result = append(result, reversedNodes)
} else {
result = append(result, curNodes)
}
curCount = nextCount
nextCount = 0
curLevel++
isAscending = !isAscending
}
return result
}
|