|
| 1 | +// Time: O(n + l) |
| 2 | +// Space: O(h + l) |
| 3 | + |
| 4 | +/** |
| 5 | + * Definition for singly-linked list. |
| 6 | + * struct ListNode { |
| 7 | + * int val; |
| 8 | + * ListNode *next; |
| 9 | + * ListNode(int x) : val(x), next(NULL) {} |
| 10 | + * }; |
| 11 | + */ |
| 12 | +/** |
| 13 | + * Definition for a binary tree node. |
| 14 | + * struct TreeNode { |
| 15 | + * int val; |
| 16 | + * TreeNode *left; |
| 17 | + * TreeNode *right; |
| 18 | + * TreeNode(int x) : val(x), left(NULL), right(NULL) {} |
| 19 | + * }; |
| 20 | + */ |
| 21 | + |
| 22 | +// kmp solution |
| 23 | +class Solution { |
| 24 | +public: |
| 25 | + bool isSubPath(ListNode* head, TreeNode* root) { |
| 26 | + if (!head) { |
| 27 | + return true; |
| 28 | + } |
| 29 | + const auto& [pattern, prefix] = getPrefix(head); |
| 30 | + return dfs(pattern, prefix, root, -1); |
| 31 | + } |
| 32 | + |
| 33 | +private: |
| 34 | + bool dfs(const vector<int>& pattern, |
| 35 | + const vector<int>& prefix, |
| 36 | + TreeNode *root, int j) { |
| 37 | + if (!root) { |
| 38 | + return false; |
| 39 | + } |
| 40 | + while (j + 1 && pattern[j + 1] != root->val) { |
| 41 | + j = prefix[j]; |
| 42 | + } |
| 43 | + if (pattern[j + 1] == root->val) { |
| 44 | + ++j; |
| 45 | + } |
| 46 | + if (j + 1 == pattern.size()) { |
| 47 | + return true; |
| 48 | + } |
| 49 | + return dfs(pattern, prefix, root->left, j) || |
| 50 | + dfs(pattern, prefix, root->right, j); |
| 51 | + } |
| 52 | + |
| 53 | + pair<vector<int>, vector<int>> getPrefix(ListNode *head) { |
| 54 | + vector<int> pattern = {head->val}; |
| 55 | + vector<int> prefix = {-1}; |
| 56 | + int j = -1; |
| 57 | + auto node = head->next; |
| 58 | + while (node) { |
| 59 | + while (j + 1 && pattern[j + 1] != node->val) { |
| 60 | + j = prefix[j]; |
| 61 | + } |
| 62 | + if (pattern[j + 1] == node->val) { |
| 63 | + ++j; |
| 64 | + } |
| 65 | + pattern.emplace_back(node->val); |
| 66 | + prefix.emplace_back(j); |
| 67 | + node = node->next; |
| 68 | + } |
| 69 | + return {pattern, prefix}; |
| 70 | + } |
| 71 | +}; |
| 72 | + |
| 73 | +// Time: O(n * min(h, l)) |
| 74 | +// Space: O(h) |
| 75 | +// dfs solution |
| 76 | +class Solution2 { |
| 77 | +public: |
| 78 | + bool isSubPath(ListNode* head, TreeNode* root) { |
| 79 | + if (!head) { |
| 80 | + return true; |
| 81 | + } |
| 82 | + if (!root) { |
| 83 | + return false; |
| 84 | + } |
| 85 | + return dfs(head, root) || |
| 86 | + isSubPath(head, root->left) || |
| 87 | + isSubPath(head, root->right); |
| 88 | + } |
| 89 | + |
| 90 | +private: |
| 91 | + bool dfs(ListNode *head, TreeNode *root) { |
| 92 | + if (!head) { |
| 93 | + return true; |
| 94 | + } |
| 95 | + if (!root) { |
| 96 | + return false; |
| 97 | + } |
| 98 | + return root->val == head->val && |
| 99 | + (dfs(head->next, root->left) || |
| 100 | + dfs(head->next, root->right)); |
| 101 | + } |
| 102 | +}; |
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