linux-安全管理-文件系统安全

class TreeNode:
    def __init__(self, weight, left = None, right = None):
        self.weight = weight
        self.left = left
        self.right = right
        self.height = 1
    #   递归计算节点的高度
    def get_height(self):
        if not self.left and not self.right:
            return 1
        left_height = self.left.get_height() if self.left else 0
        right_height = self.right.get_height() if self.right else 0
        return 1 + max(left_height, right_height)
    #   重写方法，基于节点的权重进行比较。这解决了 heapq 无法比较 Node 实例的问题
    def __lt__(self, other):
        return self.weight < other.weight
import heapq
class Solution:
    def build_huffman_tree(self, weights):
        if not weights:
            return 
        # 初始化最小堆，创建叶子节点
        heap = [TreeNode(weight) for weight in weights]
        heapq.heapify(heap)
        while len(heap) > 1:
            # 取出两个权值最小的节点
            left = heapq.heappop(heap)
            right = heapq.heappop(heap)
            # 如果权值相同，按照高度进行处理
            if left.weight == right.weight:
                left_height = left.get_height()
                right_height = right.get_height()
                # 确保左子树高度 <= 右子树高度
                if left_height > right_height:
                    left, right = right, left
            # 创建新的父节点，权值为左右节点权值之和
            merged_node = TreeNode(left.weight + right.weight, left, right)
            heapq.heappush(heap, merged_node)
        # 返回堆中唯一的节点，即哈夫曼树的根节点
        return heap[0]
    #   栈迭代法进行中序遍历
    def midorder_traversal(self, root:TreeNode):
        res = []
        if root is None:
            return
        stack = [root]
        while stack:
            node = stack.pop()
            if node != None:
                #   按照顺序压栈：右节点、中节点、左节点
                if node.right:
                    stack.append(node.right)
                stack.append(node)
                #   中节点访问过，但是还没有处理，加入空节点做为标记。
                stack.append(None)
                if node.left:
                    stack.append(node.left)
            #  只有遇到空节点的时候，才将下一个节点放进结果集
            else:
                #   按照左节点、中节点、右节点出栈
                node = stack.pop()
                res.append(node.weight)
        return res
    #   递归法进行中序遍历
    def midorder_traversal_I(self, root:TreeNode):
        res = []
        def dfs(node):
            if node is None:
                return
            dfs(node.left)
            res.append(node.weight)
            dfs(node.right)
        dfs(root)
        return res
    #   递归法简化写法
    def midorder_traversal_II(self, root:TreeNode):
        if root is None:
            return []
        return self.midorder_traversal_II(root.left) + [root.weight] + self.midorder_traversal_II(root.right)

if __name__=='__main__':
    weights = list(map(int, input().split(',')))
    solution = Solution()
    huffman_tree = solution.build_huffman_tree(weights)
    result = solution.midorder_traversal(huffman_tree)
    print(result)

知识点：哈夫曼树、堆、二叉树的中序遍历、贪心

结语：越简单的题目解法应该越多，请路过大神留下新的思路供本小白学习一下，打开思路