Leetcode 3372. Maximize the Number of Target Nodes After Connecting Trees I

• Leetcode 3372. Maximize the Number of Target Nodes After Connecting Trees I
  • 1. 解题思路
  • 2. 代码实现

• 题目链接：3372. Maximize the Number of Target Nodes After Connecting Trees I

1. 解题思路

这一题事实证明还是想得太复杂了，本来想的还是平衡和二叉树那个动态调整平衡的思路，想着怎么去优化求解最大深度为$k$的子树，结果后来发现只要最暴力的树的遍历即可……

完全进行树的遍历的话，最大的时间复杂度就是$O(N^2)$，勉强还是在允许的时间复杂度的范围内的，因此就可以省掉很多优化的思路，直接暴力求解第一个树当中所有节点作为根节点时深度不高于$k$的节点个数，然后考察第二个树当中所有节点作为根节点时深度不高于$k-1$时的节点个数，然后两者相加即可。

2. 代码实现

给出python代码实现如下：

class Solution:
    def maxTargetNodes(self, edges1: List[List[int]], edges2: List[List[int]], k: int) -> List[int]:
        graph1 = defaultdict(list)
        for u, v in edges1:
            graph1[u].append(v)
            graph1[v].append(u)

        graph2 = defaultdict(list)
        for u, v in edges2:
            graph2[u].append(v)
            graph2[v].append(u)
            
        n, m = len(edges1) + 1, len(edges2) + 1
        
        def dfs1(u, p, k):
            if k < 0:
                return 0
            elif k == 0 or len(graph1[u]) == 0:
                return 1
            elif p != -1 and len(graph1[u]) == 1:
                return 1
            return 1 + sum(dfs1(v, u, k-1) for v in graph1[u] if v != p)
        
        def dfs2(u, p, k):
            if k < 0:
                return 0
            elif k == 0 or len(graph2[u]) == 0:
                return 1
            elif p != -1 and len(graph2[u]) == 1:
                return 1
            return 1 + sum(dfs2(v, u, k-1) for v in graph2[u] if v != p)
        
        s1 = [dfs1(u, -1, k) for u in range(n)]
        s2 = [dfs2(u, -1, k-1) for u in range(m)]
        s = max(s2)
        return [x+s for x in s1] 

提交代码评测得到：耗时5515ms，占用内存19.4MB。