力扣103 二叉樹的鋸齒形遍歷

題目描述

給定乙個二叉樹，返回其節點值的鋸齒形層次遍歷。（即先從左往右，再從右往左進行下一層遍歷，以此類推，層與層之間交替進行）。

分析思路

1.雙端佇列

def zigzaglevelorder(self, root: treenode) -> list[list[int]]:

if not root:
return 
q=collections.deque()
mark=0
res=
while q:
level=
if mark%2==0:
for _ in range(len(q)):
node=q.popleft()
else:
for _ in range(len(q)):
node=q.pop()
mark+=1
return res

python:

def print( proot):

# write code here
res = 
if proot == none:
return res
stack1 = 
stack2 = 
level = 
while stack1 or stack2:
while stack1:
cur = stack1[-1]
if cur.left != none:
if cur.right != none:
stack1.pop()
if level:
level = 
while stack2:
cur = stack2[-1]
if cur.right != none:
if cur.left != none:
stack2.pop()
if level:
level = 
return res

vector> print(treenode* proot) 

if(tmp.size())
while(!stack2.empty())
if(tmp.size())
}return res;
}

完整例子如下

def print(proot):

# write code here
res = 
if proot == none:
return res
stack1 = 
stack2 = 
while stack1 or stack2:
while stack1:
cur = stack1[-1]
if cur.left != none:
if cur.right != none:
stack1.pop()
while stack2:
cur = stack2[-1]
if cur.right != none:
if cur.left != none:
stack2.pop()
return res
class treenode:
def __init__(self, x):
self.val = x
self.left = none
self.right = none
def main():
t1 = treenode(1)
t2 = treenode(2)
t3 = treenode(3)
t4 = treenode(4)
t5 = treenode(5)
t6 = treenode(6)
t7 = treenode(7)
t1.left = t2
t1.right = t3
t2.left = t4
t2.right = t5
t3.left = t6
t4.right = t7
print(print(t1))
if __name__ == "__main__":
main()

103 二叉樹的鋸齒形層次遍歷

本題跟102很像，就是將二叉樹的偶數層逆序輸出 我是直接將上一題的結果的偶數層reverse一下 definition for a binary tree node.class treenode def init self,x self.val x self.left none self.right...

103 二叉樹的鋸齒形層次遍歷

給定乙個二叉樹，返回其節點值的鋸齒形層次遍歷。即先從左往右，再從右往左進行下一層遍歷，以此類推，層與層之間交替進行 例如 給定二叉樹 3,9,20,null,null,15,7 3 9 20 15 7返回鋸齒形層次遍歷如下 3 20,9 15,7 看到本題目，首先想到的是使用佇列或棧，然而簡單使用這...

103 二叉樹的鋸齒形層次遍歷

難度 中等 題目描述 思路總結 層次遍歷 insert頭 題解一 definition for a binary tree node.class treenode def init self,x self.val x self.left none self.right none from colle...