leetcode 2385. Amount of Time for Binary Tree to Be Infected (python)

·  阅读 792

描述

You are given the root of a binary tree with unique values, and an integer start. At minute 0, an infection starts from the node with value start. Each minute, a node becomes infected if:

  • The node is currently uninfected.
  • The node is adjacent to an infected node.

Return the number of minutes needed for the entire tree to be infected.

Example 1:

Input: root = [1,5,3,null,4,10,6,9,2], start = 3
Output: 4
Explanation: The following nodes are infected during:
- Minute 0: Node 3
- Minute 1: Nodes 1, 10 and 6
- Minute 2: Node 5
- Minute 3: Node 4
- Minute 4: Nodes 9 and 2
It takes 4 minutes for the whole tree to be infected so we return 4.

复制代码

Example 2:

Input: root = [1], start = 1
Output: 0
Explanation: At minute 0, the only node in the tree is infected so we return 0.
复制代码

Note:

The number of nodes in the tree is in the range [1, 10^5].
1 <= Node.val <= 10^5
Each node has a unique value.
A node with a value of start exists in the tree.
复制代码

解析

根据题意,给定具有唯一值的二叉树的根 root 和一个整数 start 。 在第 0 分钟,感染从值为 start 的节点开始。 每分钟,如果出现以下情况,节点就会被感染:

  • 该节点当前未受感染。
  • 该节点与受感染的节点相邻。

返回整个树被感染所需的分钟数。

这个题很明显整体需要 BFS 的方式进行解题,但是关键在于这是一颗树,我们要使用两个字典 parent 和 child 来记录每个节点的父节点和子节点有哪些,这个过程需要最基础的 DFS 进行解决,然后我们得到 parent 和 child 之后,就可以按照 BFS 的思路从 start 开始模拟病毒扩散的效果来解题,使用 result 来记录扩散时间即可。虽然代码量上比较大,但是很好理解,上半部分就是 DFS 过程,下半部分就是 BFS 过程。当然代码还可以优化,有兴趣的同学可以尝试一下。

时间复杂度为 O(N) ,空间复杂度为 O(N) 。

解答

class Solution(object):
    def __init__(self):
        self.parent = {}
        self.child = collections.defaultdict(list)
    def amountOfTime(self, root, start):
        def dfs(root):
            if not root:
                return
            if root.left:
                self.parent[root.left.val] = root
                self.child[root.val].append(root.left)
                dfs(root.left)
            if root.right:
                self.parent[root.right.val] = root
                self.child[root.val].append(root.right)
                dfs(root.right)
        dfs(root)
        result = 0
        if start == root.val and not self.parent and not self.child:
            return result
        stack = [start]
        visited = set()
        while stack:
            for _ in range(len(stack)):
                node = stack.pop(0)
                visited.add(node)
                if node in self.parent and self.parent[node].val not in visited:
                    stack.append(self.parent[node].val)
                for c in self.child[node]:
                    if c.val not in visited:
                        stack.append(c.val)
            if stack:
                result += 1
        return result
复制代码

运行结果

80 / 80 test cases passed.
Status: Accepted
Runtime: 2872 ms
Memory Usage: 176.1 MB
复制代码

原题链接

leetcode.com/contest/wee…

您的支持是我最大的动力

分类:
后端
收藏成功!
已添加到「」, 点击更改