Javascript 🎉🎉🎉
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```javascript []\n/**\n * @param {number[][]} edges1\n * @param {number[][]} edges2\n * @param {number} k\n * @return {number[]}\n */\nvar maxTargetNodes = function(edges1, edges2, k) {\n    const n = edges1.length + 1;\n    const m = edges2.length + 1;\n\n    const adj1 = Array.from({ length: n }, () => []);\n    const adj2 = Array.from({ length: m }, () => []);\n\n    for (const [u, v] of edges1) {\n      adj1[u].push(v);\n      adj1[v].push(u);\n    }\n    for (const [u, v] of edges2) {\n      adj2[u].push(v);\n      adj2[v].push(u);\n    } \n    const good1 = new Array(n).fill(0);\n    const good2 = new Array(m).fill(0);\n\n    const dfs = (node, parent, distance, root, k, good, adj) => {\n      if (distance >= k) return;\n      good[root]++;\n      for (const neighbor of adj[node]) {\n        if (neighbor !== parent) {\n          dfs(neighbor, node, distance + 1, root, k, good, adj);\n        }\n      }\n    };\n    for (let i = 0; i < n; i++) {\n      dfs(i, -1, 0, i, k + 1, good1, adj1);\n    }\n\n    for (let i = 0; i < m; i++) {\n      dfs(i, -1, 0, i, k, good2, adj2);\n    }\n\n    const mx = Math.max(...good2);\n\n    return good1.map(value => value + mx);\n\n};\n```
['JavaScript']