12 Graph Valid Tree — Interview English Script (C++)
Source aligned with: docs/15_Graphs/12_Graph_Valid_Tree.md
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1) 30-second problem restatement script
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me restate graph valid tree. | 我先重述 Graph Valid Tree。 | Restatement |
| We have n nodes and undirected edges. | 題目給 n 個節點與無向邊。 | Restatement |
| We need to decide whether the graph is a valid tree. | 要判斷這張圖是否為有效樹。 | Restatement |
| A valid tree must be connected and acyclic. | 有效樹必須連通且無環。 | Restatement |
| A necessary property is edge count equals n minus one. | 必要條件是邊數要等於 n-1。 | Restatement |
| I will use Union Find plus edge-count check. | 我會用 Union Find 搭配邊數檢查。 | Restatement |
2) Clarifying questions (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Are node labels from zero to n minus one? | 節點編號是否為 0 到 n-1？ | Clarify |
| Is graph undirected for every edge pair? | 每條邊是否都是無向？ | Clarify |
| Can there be duplicate edges or self-loops? | 是否可能有重複邊或自迴圈？ | Clarify |
| Should I return boolean only? | 是否只需回傳布林值？ | Clarify |
| Is early check edges size not equal n minus one acceptable? | 可否先做 edges 數量不等於 n-1 的早退檢查？ | Clarify |
3) Approach discussion
Brute force (3 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| DFS or BFS can test connectivity and cycle. | DFS/BFS 可以檢查連通與環。 | Approach |
| But we still need careful parent tracking for cycle in undirected graph. | 但無向圖判環仍需小心父節點追蹤。 | Approach |
| Union Find gives concise checks for both conditions. | Union Find 可精簡地同時處理兩條件。 | Approach |
Optimized approach (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| First check edges size equals n minus one. | 先檢查邊數是否為 n-1。 | Approach |
| If not equal, graph cannot be a tree. | 若不相等，圖不可能是樹。 | Approach |
| Then run Union Find over all edges. | 接著用 Union Find 掃所有邊。 | Approach |
| If any edge connects nodes already in same set, cycle exists so false. | 若任何邊連到同集合，表示有環，回 false。 | Approach |
| If all unions succeed under n minus one edges, graph is connected and valid. | 在 n-1 邊下若都可合併成功，圖即連通且有效。 | Approach |
4) Coding-and-speaking script (line-by-line, in coding order)
| English line | Traditional Chinese meaning (short) | Interview stage |
| I first check if edges size is not n minus one, then return false. | 我先檢查 edges 數量若不是 n-1 就回 false。 | Coding |
| I initialize parent array parent[i]=i. | 我初始化 parent 陣列 parent[i]=i。 | Coding |
| I may keep components count starting at n. | 我可以維護 components 從 n 開始。 | Coding |
| For each edge u v, compute roots ru and rv. | 對每條邊 u,v 計算根 ru、rv。 | Coding |
| If ru equals rv, this edge forms cycle so return false. | 若 ru==rv，這條邊成環，回 false。 | Coding |
| Otherwise union ru and rv. | 否則合併 ru 與 rv。 | Coding |
| Optionally decrement components after each union. | 可選擇在每次 union 後遞減 components。 | Coding |
| Continue processing all edges. | 持續處理直到所有邊完成。 | Coding |
| If no cycle found, return true. | 若未發現環，就回 true。 | Coding |
| This works because edge-count precheck guarantees connectivity. | 這可行是因為邊數前置檢查已保障連通性條件。 | Coding |
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me dry-run n five with edges [0,1],[0,2],[0,3],[1,4]. | 我手跑 n=5，邊為 [0,1],[0,2],[0,3],[1,4]。 | Dry-run |
| Edge count is four, equal to n minus one. | 邊數是 4，等於 n-1。 | Dry-run |
| Union zero one succeeds. | 合併 0 與 1 成功。 | Dry-run |
| Union zero two and zero three both succeed. | 合併 0-2 與 0-3 都成功。 | Dry-run |
| Union one four also succeeds without same-root conflict. | 合併 1-4 也成功，沒有同根衝突。 | Dry-run |
| No cycle is detected in all edges. | 所有邊都未偵測到環。 | Dry-run |
| Return true for valid tree. | 回傳 true，表示有效樹。 | Dry-run |
6) Edge/corner test script (at least 4 cases)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Case one: edges less than n minus one means disconnected, return false. | 案例一：邊少於 n-1 代表不連通，回 false。 | Edge test |
| Case two: edges greater than n minus one means cycle guaranteed, return false. | 案例二：邊多於 n-1 保證有環，回 false。 | Edge test |
| Case three: exactly n minus one with cycle still false. | 案例三：雖是 n-1 但有環仍回 false。 | Edge test |
| Case four: n one with no edge should be true. | 案例四：n=1 且無邊應為 true。 | Edge test |
| Case five: self-loop edge should fail cycle check. | 案例五：自迴圈邊應觸發成環失敗。 | Edge test |
7) Complexity script
Short version (2 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Time complexity is O(N alpha N). | 時間複雜度是 O(N·α(N))。 | Complexity |
| Space complexity is O(N). | 空間複雜度是 O(N)。 | Complexity |
Full version (4 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| We process about n minus one edges after precheck. | 前置檢查後大約處理 n-1 條邊。 | Complexity |
| Each find or union is near constant with path compression. | 路徑壓縮下每次 find/union 近乎常數。 | Complexity |
| Total runtime is O(N alpha N). | 總時間是 O(N·α(N))。 | Complexity |
| Parent array uses O(N) memory. | parent 陣列使用 O(N) 記憶體。 | Complexity |
8) If stuck rescue lines (10 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me separate tree conditions clearly. | 我先把樹的條件分開講清楚。 | If stuck |
| Condition one: exactly n minus one edges. | 條件一：邊數必須是 n-1。 | If stuck |
| Condition two: no cycle. | 條件二：不可有環。 | If stuck |
| Under condition one, no cycle implies connected. | 在條件一成立下，無環就能推出連通。 | If stuck |
| So edge-count precheck simplifies final logic. | 所以邊數前檢可簡化後續邏輯。 | If stuck |
| In Union Find, same root means cycle. | Union Find 中同根代表成環。 | If stuck |
| Let me test a cycle sample quickly. | 我快速測一個成環案例。 | If stuck |
| Same-root detection should trigger false immediately. | 同根檢測應立即回 false。 | If stuck |
| Then test simple chain should return true. | 再測簡單鏈狀圖應回 true。 | If stuck |
| Great, correctness intuition is now solid. | 很好，正確性直覺已穩固。 | If stuck |
9) Final wrap-up lines (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| I solved valid tree with edge-count check plus Union Find. | 我用邊數檢查加 Union Find 解了 valid tree。 | Wrap-up |
| Edge count filters impossible cases early. | 邊數檢查可先過濾不可能案例。 | Wrap-up |
| Same-root union attempt detects cycles directly. | union 時同根可直接偵測環。 | Wrap-up |
| Complexity is near linear with O(N alpha N) time and O(N) space. | 複雜度近線性：O(N·α(N)) 時間、O(N) 空間。 | Wrap-up |
| This is a standard DSU tree-validation pattern. | 這是樹驗證題的標準 DSU 模式。 | Wrap-up |
10) Ultra-short cheat sheet (20 lines total)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Goal: check if undirected graph is a tree. | 目標：判斷無向圖是否為樹。 | Cheat sheet |
| Tree needs connected plus acyclic. | 樹需要連通且無環。 | Cheat sheet |
| Precheck edges count == n-1. | 先檢查邊數是否 n-1。 | Cheat sheet |
| If not, return false immediately. | 若否，立即回 false。 | Cheat sheet |
| Initialize Union Find parent array. | 初始化 Union Find parent。 | Cheat sheet |
| Process each edge u v. | 處理每條邊 u,v。 | Cheat sheet |
| Find roots ru and rv. | 查根 ru、rv。 | Cheat sheet |
| If ru==rv, cycle -> false. | 若 ru==rv，有環 -> false。 | Cheat sheet |
| Else union roots. | 否則合併兩根。 | Cheat sheet |
| Continue all edges. | 持續處理所有邊。 | Cheat sheet |
| No cycle found => true. | 未發現環 => true。 | Cheat sheet |
| n=1 no edge => true. | n=1 無邊 => true。 | Cheat sheet |
| Too few edges => disconnected false. | 邊太少 => 不連通 false。 | Cheat sheet |
| Too many edges => cycle false. | 邊太多 => 有環 false。 | Cheat sheet |
| Time O(N alpha N). | 時間 O(N·α(N))。 | Cheat sheet |
| Space O(N). | 空間 O(N)。 | Cheat sheet |
| Common bug: forgetting edge-count precheck. | 常見錯誤：忘記邊數前檢。 | Cheat sheet |
| Common bug: wrong index bounds in parent. | 常見錯誤：parent 索引範圍錯。 | Cheat sheet |
| DFS/BFS can also solve. | DFS/BFS 也能解。 | Cheat sheet |
| Explain two conditions clearly to interviewer. | 跟面試官清楚說兩個條件。 | Cheat sheet |
Quality check
- Consistency with source solution: ✅ Edge-count precheck + DSU cycle detection aligned.
- No hallucinated constraints: ✅ Tree definition and return semantics preserved.
- Language simplicity: ✅ Short and practical interview speech.