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| 1 | +// Time: O(n) |
| 2 | +// Space: O(w) |
| 3 | + |
| 4 | +/** |
| 5 | + * Definition for a binary tree node. |
| 6 | + * struct TreeNode { |
| 7 | + * int val; |
| 8 | + * TreeNode *left; |
| 9 | + * TreeNode *right; |
| 10 | + * TreeNode() : val(0), left(nullptr), right(nullptr) {} |
| 11 | + * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} |
| 12 | + * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} |
| 13 | + * }; |
| 14 | + */ |
| 15 | +// bfs solution |
| 16 | +class Solution { |
| 17 | +public: |
| 18 | + bool isValidSequence(TreeNode* root, vector<int>& arr) { |
| 19 | + vector<TreeNode *> q = {root}; |
| 20 | + for (int depth = 0; depth < arr.size(); ++depth) { |
| 21 | + vector<TreeNode *> new_q; |
| 22 | + while (!q.empty()) { |
| 23 | + const auto node = q.back(); q.pop_back(); |
| 24 | + if (!node || node->val != arr[depth]) { |
| 25 | + continue; |
| 26 | + } |
| 27 | + if (depth + 1 == arr.size() && node->left == node->right) { |
| 28 | + return true; |
| 29 | + } |
| 30 | + new_q.emplace_back(node->left); |
| 31 | + new_q.emplace_back(node->right); |
| 32 | + } |
| 33 | + q = move(new_q); |
| 34 | + } |
| 35 | + return false; |
| 36 | + } |
| 37 | +}; |
| 38 | + |
| 39 | +// Time: O(n) |
| 40 | +// Space: O(h) |
| 41 | +// dfs solution with stack |
| 42 | +class Solution2 { |
| 43 | +public: |
| 44 | + bool isValidSequence(TreeNode* root, vector<int>& arr) { |
| 45 | + vector<pair<TreeNode *, int>> s = {{root, 0}}; |
| 46 | + while (!s.empty()) { |
| 47 | + const auto [node, depth] = s.back(); s.pop_back(); |
| 48 | + if (!node || depth == arr.size() || node->val != arr[depth]) { |
| 49 | + continue; |
| 50 | + } |
| 51 | + if (depth + 1 == arr.size() && node->left == node->right) { |
| 52 | + return true; |
| 53 | + } |
| 54 | + s.emplace_back(node->right, depth + 1); |
| 55 | + s.emplace_back(node->left, depth + 1); |
| 56 | + } |
| 57 | + return false; |
| 58 | + } |
| 59 | +}; |
| 60 | + |
| 61 | +// Time: O(n) |
| 62 | +// Space: O(h) |
| 63 | +// dfs solution with recursion |
| 64 | +class Solution3 { |
| 65 | +public: |
| 66 | + bool isValidSequence(TreeNode* root, vector<int>& arr) { |
| 67 | + return dfs(root, arr, 0); |
| 68 | + } |
| 69 | + |
| 70 | +private: |
| 71 | + bool dfs(TreeNode *node, const vector<int>& arr, int depth) { |
| 72 | + if (!node || depth == arr.size() || node->val != arr[depth]) { |
| 73 | + return false; |
| 74 | + } |
| 75 | + if (depth + 1 == arr.size() && node->left == node->right) { |
| 76 | + return true; |
| 77 | + } |
| 78 | + return dfs(node->left, arr, depth + 1) || dfs(node->right, arr, depth + 1); |
| 79 | + } |
| 80 | +}; |
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