Java Solution Using DFS
Approach\nStep 1: Represent the Trees\nUse adjacency lists to represent the graphs tree1 and tree2 for efficient traversal.\n\nStep 2: Compute Reachable Nodes f
# Approach\nStep 1: Represent the Trees\nUse adjacency lists to represent the graphs tree1 and tree2 for efficient traversal.\n\nStep 2: Compute Reachable Nodes for tree1\nFor each node in tree1, perform a DFS (or BFS) up to a depth of k. During the DFS:\nTrack visited nodes to avoid cycles.\nCount the number of nodes reachable within k moves.\n\nStep 3: Compute the Maximum Reachable Nodes for tree2\nFor all nodes in tree2, perform a DFS (or BFS) up to a depth of k-1. During the DFS:\nTrack visited nodes to avoid cycles.\nFind the maximum number of reachable nodes for any starting node in tree2.\n\nStep 4: Combine Results\nFor each node in tree1, add the precomputed reachable nodes (from Step 2) to the maximum reachable nodes in tree2 (from Step 3).\n\nStep 5: Return the Results\nReturn an array where each entry corresponds to the sum of reachable nodes in tree1 and the maximum reachable nodes in tree2.\n\n# Complexity\n- Time complexity:\nO((n + m)*k);\n\n- Space complexity:\nO(n + m + k);\n\n# Code\n```java []\nimport java.util.*;\n\nclass Solution {\n    public int[] maxTargetNodes(int[][] edges1, int[][] edges2, int k) {\n        int n = edges1.length + 1;\n        int m = edges2.length + 1;\n        int[] answer = new int[n];\n        ArrayList<Integer>[] tree1 = new ArrayList[n];\n        ArrayList<Integer>[] tree2 = new ArrayList[m];\n\n        for (int i = 0; i < n; i++) tree1[i] = new ArrayList<>();\n        for (int i = 0; i < m; i++) tree2[i] = new ArrayList<>();\n\n        for (int[] edge : edges1) {\n            tree1[edge[0]].add(edge[1]);\n            tree1[edge[1]].add(edge[0]);\n        }\n\n        for (int[] edge : edges2) {\n            tree2[edge[0]].add(edge[1]);\n            tree2[edge[1]].add(edge[0]);\n        }\n\n        int[] temp = new int[n];\n        for (int i = 0; i < n; i++) {\n            boolean[] visited = new boolean[n];\n            visited[i] = true;\n            temp[i] = dfs(tree1, i, k, visited);\n        }\n\n        int max = 0;\n        for (int i = 0; i < m; i++) {\n            boolean[] visited = new boolean[m];\n            visited[i] = true;\n            max = Math.max(max, dfs(tree2, i, k - 1, visited));\n        }\n\n        for (int i = 0; i < n; i++) {\n            answer[i] = temp[i] + max;\n        }\n\n        return answer;\n    }\n\n    public int dfs(ArrayList<Integer>[] graph, int src, int k, boolean[] visited) {\n        if (k < 0) return 0;\n        int cnt = 1;\n        for (int nbr : graph[src]) {\n            if (!visited[nbr]) {\n                visited[nbr] = true;\n                cnt += dfs(graph, nbr, k - 1, visited);\n            }\n        }\n        return cnt;\n    }\n}\n\n```
['Depth-First Search', 'Java']