剑指 Offer 32 - III. 从上到下打印二叉树 III
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],
3
/ \
9 20
/ \
15 7
返回其层次遍历结果：
[ [3], [20,9], [15,7] ]
提示：
节点总数 <= 1000原站题解
python3 解法, 执行用时: 40 ms, 内存消耗: 15.2 MB, 提交时间: 2022-11-14 10:50:28
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root: return []
res, queue = [], collections.deque()
queue.append(root)
while queue:
tmp = []
for _ in range(len(queue)):
node = queue.popleft()
tmp.append(node.val)
if node.left: queue.append(node.left)
if node.right: queue.append(node.right)
res.append(tmp[::-1] if len(res) % 2 else tmp)
return res
python3 解法, 执行用时: 48 ms, 内存消耗: 15.3 MB, 提交时间: 2022-11-14 10:50:07
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root: return []
res, deque = [], collections.deque()
deque.append(root)
while deque:
tmp = []
# 打印奇数层
for _ in range(len(deque)):
# 从左向右打印
node = deque.popleft()
tmp.append(node.val)
# 先左后右加入下层节点
if node.left: deque.append(node.left)
if node.right: deque.append(node.right)
res.append(tmp)
if not deque: break # 若为空则提前跳出
# 打印偶数层
tmp = []
for _ in range(len(deque)):
# 从右向左打印
node = deque.pop()
tmp.append(node.val)
# 先右后左加入下层节点
if node.right: deque.appendleft(node.right)
if node.left: deque.appendleft(node.left)
res.append(tmp)
return res
python3 解法, 执行用时: 40 ms, 内存消耗: 15.2 MB, 提交时间: 2022-11-14 10:49:37
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root: return []
res, deque = [], collections.deque([root])
while deque:
tmp = collections.deque()
for _ in range(len(deque)):
node = deque.popleft()
if len(res) % 2: tmp.appendleft(node.val) # 偶数层 -> 队列头部
else: tmp.append(node.val) # 奇数层 -> 队列尾部
if node.left: deque.append(node.left)
if node.right: deque.append(node.right)
res.append(list(tmp))
return res
golang 解法, 执行用时: 0 ms, 内存消耗: 2.6 MB, 提交时间: 2022-11-14 10:38:27
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func levelOrder(root *TreeNode) (ans [][]int) {
if root == nil {
return
}
queue := []*TreeNode{root}
for level := 0; len(queue) > 0; level++ {
vals := []int{}
q := queue
queue = nil
for _, node := range q {
vals = append(vals, node.Val)
if node.Left != nil {
queue = append(queue, node.Left)
}
if node.Right != nil {
queue = append(queue, node.Right)
}
}
// 本质上和层序遍历一样，我们只需要把奇数层的元素翻转即可
if level%2 == 1 {
for i, n := 0, len(vals); i < n/2; i++ {
vals[i], vals[n-1-i] = vals[n-1-i], vals[i]
}
}
ans = append(ans, vals)
}
return
}