請實現一個函數按照之字形順序打印二叉樹,即第一行按照從左到右的順序打印,第二層按照從右到左的順序打印,第三行再按照從左到右的順序打印,其他行以此類推。
例如:
給定二叉樹: [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
返回其層次遍歷結果:
[
[3],
[20,9],
[15,7]
]
提示:
節點總數 <= 1000
來源:力扣(LeetCode)
鏈接:https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof
著作權歸領釦網絡所有。商業轉載請聯繫官方授權,非商業轉載請註明出處。
思路1:dfs層序遍歷
此題與LeetCodeEasy-【面試題32 - II. 從上到下打印二叉樹 II】基本類似,唯一的區別就是隔一層需要反轉以下,於是我們可以將得到的值利用list的性質反轉即可。
# 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 root == None:
return []
# dfs
father = [root]
child = []
ans = []
layer = []
f = 1
while father != [] or child != []:
if father == []:
if f == 1:
ans.append(layer)
else:
ans.append(layer[::-1])
f = -1*f
layer = []
t = father
father = child
child = t
node = father.pop(0)
layer.append(node.val)
if node.left != None:
child.append(node.left)
if node.right != None:
child.append(node.right)
if f == 1:
ans.append(layer)
else:
ans.append(layer[::-1])
return ans
思路2:利用前一題思路改編
# 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 root == None:
return []
# dfs
stack = collections.deque()
ans = []
stack.append(root)
f = 1
while stack:
t = []
for _ in range(len(stack)):
root = stack.popleft()
t.append(root.val)
if root.left:
stack.append(root.left)
if root.right:
stack.append(root.right)
if f == 1:
ans.append(t)
else:
ans.append(t[::-1])
f *= -1
return ans
思路3:官方題解
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) # 偶數層 -> 隊列頭部
else: tmp.append(node.val) # 奇數層 -> 隊列尾部
if node.left: deque.append(node.left)
if node.right: deque.append(node.right)
res.append(list(tmp))
return res
作者:jyd
鏈接:https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof/solution/mian-shi-ti-32-iii-cong-shang-dao-xia-da-yin-er--3/
來源:力扣(LeetCode)
著作權歸作者所有。商業轉載請聯繫作者獲得授權,非商業轉載請註明出處。