LeetCode刷题第八天

LeetCode刷题记录
1.顺时针打印矩阵
输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。
示例：输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]

题解

class Solution:
    def spiralOrder(self, matrix:[[int]]) -> [int]:
        if not matrix: return []
        l, r, t, b, res = 0, len(matrix[0]) - 1, 0, len(matrix) - 1, []
        while True:
            for i in range(l, r + 1): res.append(matrix[t][i]) # left to right
            t += 1
            if t > b: break
            for i in range(t, b + 1): res.append(matrix[i][r]) # top to bottom
            r -= 1
            if l > r: break
            for i in range(r, l - 1, -1): res.append(matrix[b][i]) # right to left
            b -= 1
            if t > b: break
            for i in range(b, t - 1, -1): res.append(matrix[i][l]) # bottom to top
            l += 1
            if l > r: break
        return res

2.对称的二叉树
请实现一个函数，用来判断一棵二叉树是不是对称的。如果一棵二叉树和它的镜像一样，那么它是对称的。
题解

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        def recur(L,R):
            if not L and not R: return True
            if not L or not R or L.val != R.val:return False
            return recur(L.left, R.right) and recur(L.right, R.left)

        return recur(root.left, root.right) if root else True

3.包含min函数的栈
定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。
示例：
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.min(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.min(); --> 返回 -2.

题解：

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.A, self.B = [], []


    def push(self, x: int) -> None:
        self.A.append(x)
        if not self.B or self.B[-1] >= x:
            self.B.append(x)

    def pop(self) -> None:
        if self.A.pop() == self.B[-1]:
            self.B.pop()

    def top(self) -> int:
        return self.A[-1]

    def min(self) -> int:
        return self.B[-1]


# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.min()

4.打印二叉树
从上到下打印出二叉树的每个节点，同一层的节点按照从左到右的顺序打印。

题解：

#1
# 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[int]:
        if not root: return []
        res, queue = [], collections.deque()
        queue.append(root)
        while queue:
            node = queue.popleft()
            res.append(node.val)
            if node.left: queue.append(node.left)
            if node.right: queue.append(node.right)
        return res
#2
# 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[int]:
        if not root: return []
        que_node=[root]
        ans = []
        while que_node:
            cur=que_node.pop(0)
            ans.append(cur.val)
            if cur.left: que_node.append(cur.left)
            if cur.right: que_node.append(cur.right)
        return ans