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144.二叉树的前序遍历.py

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+#
+# @lc app=leetcode.cn id=144 lang=python3
+#
+# [144] 二叉树的前序遍历
+#
+# https://leetcode-cn.com/problems/binary-tree-preorder-traversal/description/
+#
+# algorithms
+# Medium (64.01%)
+# Likes:    208
+# Dislikes: 0
+# Total Accepted:    70.7K
+# Total Submissions: 109.9K
+# Testcase Example:  '[1,null,2,3]'
+#
+# 给定一个二叉树，返回它的 前序 遍历。
+# 
+# 示例:
+# 
+# 输入: [1,null,2,3]  
+# ⁠  1
+# ⁠   \
+# ⁠    2
+# ⁠   /
+# ⁠  3 
+# 
+# 输出: [1,2,3]
+# 
+# 
+# 进阶: 递归算法很简单，你可以通过迭代算法完成吗？
+# 
+#
+
+# @lc code=start
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def preorderTraversal(self, root: TreeNode) -> List[int]:
+        # 栈
+        res = []
+        if not root:
+            return res
+        stack = [root]
+        while stack:
+            cur = stack.pop()
+            res.append(cur.val)
+            if cur.right:
+                stack.append(cur.right)
+            if cur.left:
+                stack.append(cur.left)
+        return res
+# @lc code=end
+

145.二叉树的后序遍历.py

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+#
+# @lc app=leetcode.cn id=145 lang=python3
+#
+# [145] 二叉树的后序遍历
+#
+# https://leetcode-cn.com/problems/binary-tree-postorder-traversal/description/
+#
+# algorithms
+# Hard (70.07%)
+# Likes:    228
+# Dislikes: 0
+# Total Accepted:    52.6K
+# Total Submissions: 74.8K
+# Testcase Example:  '[1,null,2,3]'
+#
+# 给定一个二叉树，返回它的 后序 遍历。
+# 
+# 示例:
+# 
+# 输入: [1,null,2,3]  
+# ⁠  1
+# ⁠   \
+# ⁠    2
+# ⁠   /
+# ⁠  3 
+# 
+# 输出: [3,2,1]
+# 
+# 进阶: 递归算法很简单，你可以通过迭代算法完成吗？
+# 
+#
+
+# @lc code=start
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def postorderTraversal(self, root: TreeNode) -> List[int]:
+        # 栈：前序遍历左右互换后逆序输出
+        res = []
+        if not root:
+            return res
+        stack = [root]
+        while stack:
+            cur = stack.pop()
+            res.append(cur.val)
+            if cur.left:
+                stack.append(cur.left)
+            if cur.right:
+                stack.append(cur.right)
+        return res[::-1]
+# @lc code=end
+

153.寻找旋转排序数组中的最小值.py

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+#
+# @lc app=leetcode.cn id=153 lang=python3
+#
+# [153] 寻找旋转排序数组中的最小值
+#
+# https://leetcode-cn.com/problems/find-minimum-in-rotated-sorted-array/description/
+#
+# algorithms
+# Medium (50.19%)
+# Likes:    143
+# Dislikes: 0
+# Total Accepted:    33K
+# Total Submissions: 65.8K
+# Testcase Example:  '[3,4,5,1,2]'
+#
+# 假设按照升序排序的数组在预先未知的某个点上进行了旋转。
+# 
+# ( 例如，数组 [0,1,2,4,5,6,7]  可能变为 [4,5,6,7,0,1,2] )。
+# 
+# 请找出其中最小的元素。
+# 
+# 你可以假设数组中不存在重复元素。
+# 
+# 示例 1:
+# 
+# 输入: [3,4,5,1,2]
+# 输出: 1
+# 
+# 示例 2:
+# 
+# 输入: [4,5,6,7,0,1,2]
+# 输出: 0
+# 
+#
+
+# @lc code=start
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        # 二分查找
+        left, right = 0, len(nums) - 1
+        while left < right:  
+            mid = (left + right) // 2
+            if nums[mid] > nums[right]:
+                left = mid + 1
+            else:
+                right = mid
+        return nums[left]  # left == right 结束循环 min = left or right
+
+# @lc code=end
+

199.二叉树的右视图.py

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+#
+# @lc app=leetcode.cn id=199 lang=python3
+#
+# [199] 二叉树的右视图
+#
+# https://leetcode-cn.com/problems/binary-tree-right-side-view/description/
+#
+# algorithms
+# Medium (63.19%)
+# Likes:    139
+# Dislikes: 0
+# Total Accepted:    18.4K
+# Total Submissions: 29K
+# Testcase Example:  '[1,2,3,null,5,null,4]'
+#
+# 给定一棵二叉树，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。
+# 
+# 示例:
+# 
+# 输入: [1,2,3,null,5,null,4]
+# 输出: [1, 3, 4]
+# 解释:
+# 
+# ⁠  1            <---
+# ⁠/   \
+# 2     3         <---
+# ⁠\     \
+# ⁠ 5     4       <---
+# 
+# 
+#
+
+# @lc code=start
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def rightSideView(self, root: TreeNode) -> List[int]:
+        # BFS queue
+        view = []
+        if not root:
+            return view
+        level = [root]
+        while level:
+            view.append(level[-1].val)
+            for i in range(len(level)):
+                top = level.pop(0)
+                if top.left:
+                    level.append(top.left)
+                if top.right:
+                    level.append(top.right)
+        return view
+# @lc code=end
+

215.数组中的第k个最大元素.py

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+#
+# @lc app=leetcode.cn id=215 lang=python3
+#
+# [215] 数组中的第K个最大元素
+#
+# https://leetcode-cn.com/problems/kth-largest-element-in-an-array/description/
+#
+# algorithms
+# Medium (61.03%)
+# Likes:    379
+# Dislikes: 0
+# Total Accepted:    81.7K
+# Total Submissions: 133.5K
+# Testcase Example:  '[3,2,1,5,6,4]\n2'
+#
+# 在未排序的数组中找到第 k 个最大的元素。请注意，你需要找的是数组排序后的第 k 个最大的元素，而不是第 k 个不同的元素。
+# 
+# 示例 1:
+# 
+# 输入: [3,2,1,5,6,4] 和 k = 2
+# 输出: 5
+# 
+# 
+# 示例 2:
+# 
+# 输入: [3,2,3,1,2,4,5,5,6] 和 k = 4
+# 输出: 4
+# 
+# 说明: 
+# 
+# 你可以假设 k 总是有效的，且 1 ≤ k ≤ 数组的长度。
+# 
+#
+
+# @lc code=start
+import heapq
+
+class Solution:
+    def findKthLargest(self, nums: List[int], k: int) -> int:
+        # 最小堆
+        h = []
+        for i in range(k):
+            heapq.heappush(h, nums[i])
+        for j in range(k, len(nums)):
+            top = h[0]
+            if nums[j] > top:
+                heapq.heapreplace(h, nums[j])
+        return h[0]
+# @lc code=end
+

232.用栈实现队列.py

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+#
+# @lc app=leetcode.cn id=232 lang=python3
+#
+# [232] 用栈实现队列
+#
+# https://leetcode-cn.com/problems/implement-queue-using-stacks/description/
+#
+# algorithms
+# Easy (62.72%)
+# Likes:    140
+# Dislikes: 0
+# Total Accepted:    32.9K
+# Total Submissions: 52.2K
+# Testcase Example:  '["MyQueue","push","push","peek","pop","empty"]\n[[],[1],[2],[],[],[]]'
+#
+# 使用栈实现队列的下列操作：
+# 
+# 
+# push(x) -- 将一个元素放入队列的尾部。
+# pop() -- 从队列首部移除元素。
+# peek() -- 返回队列首部的元素。
+# empty() -- 返回队列是否为空。
+# 
+# 
+# 示例:
+# 
+# MyQueue queue = new MyQueue();
+# 
+# queue.push(1);
+# queue.push(2);  
+# queue.peek();  // 返回 1
+# queue.pop();   // 返回 1
+# queue.empty(); // 返回 false
+# 
+# 说明:
+# 
+# 
+# 你只能使用标准的栈操作 -- 也就是只有 push to top, peek/pop from top, size, 和 is empty
+# 操作是合法的。
+# 你所使用的语言也许不支持栈。你可以使用 list 或者 deque（双端队列）来模拟一个栈，只要是标准的栈操作即可。
+# 假设所有操作都是有效的 （例如，一个空的队列不会调用 pop 或者 peek 操作）。
+# 
+# 
+#
+
+# @lc code=start
+class MyQueue:
+
+    def __init__(self):
+        """
+        Initialize your data structure here.
+        """
+        self.in_stack, self.out_stack = [], []
+
+
+    def push(self, x: int) -> None:
+        """
+        Push element x to the back of queue.
+        """
+        self.in_stack.append(x)
+
+
+    def pop(self) -> int:
+        """
+        Removes the element from in front of queue and returns that element.
+        """
+        self.move()
+        return self.out_stack.pop()
+
+    def peek(self) -> int:
+        """
+        Get the front element.
+        """
+        self.move()
+        return self.out_stack[-1]
+
+    def empty(self) -> bool:
+        """
+        Returns whether the queue is empty.
+        """
+        return not self.in_stack and not self.out_stack
+
+    def move(self):
+        if not self.out_stack:
+            while self.in_stack:
+                self.out_stack.append(self.in_stack.pop())
+
+
+# Your MyQueue object will be instantiated and called as such:
+# obj = MyQueue()
+# obj.push(x)
+# param_2 = obj.pop()
+# param_3 = obj.peek()
+# param_4 = obj.empty()
+# @lc code=end
+