LeetCode刷题总结python3

# Definition for a binary tree node.

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

# Definition for singly-linked list.

# class ListNode:

# def __init__(self, x):

# self.val = x

# self.next = None

调用函数时，加 self.函数名

list排序 nums.sort()

pre.next = l1 if l1 else l2 # 类似C++三目表达式，l1为真则l1，否则l2

list列表使用

def convertBST(self, root: TreeNode) -> TreeNode:

stack = [(root,1)]

sum = 0

while stack:

node, cmd = stack.pop()

if node is None:

continue

if cmd == 0:

sum += node.val

node.val = sum

else:

stack.append((node.left, 1))

stack.append((node, 0))

stack.append((node.right, 1))

return root

python定义全局变量

self.sum =0

def convertBST(self, root: TreeNode) -> TreeNode:

if root == None:

return

self.val = 0

def dfs(root):

if root == None:

return

root.right = dfs(root.right)

self.val += root.val

root.val = self.val

root.left = dfs(root.left)

return root

return dfs(root)

python3交换

a, b = b, a # 类似C++ swap操作

list模拟栈

self.stackB = []

if self.stackB: # 判断是否为空

self.stackB[-1] # 取top()

self.stackB.append(x) # push操作

self.stackB.pop() # pop()操作

创建list

ary_len =len(prices)

dp= [0] * ary_len

import collections

d = collections.deque()

import collections

d = collections.deque()

d.append(1)

d.append(2)

print(d)

#输出：deque([1, 2])

d.appendleft(2)

d.clear()

x = d.pop()

x = d.popleft()

d.reverse()