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| 1 | +// Time: O(n) |
| 2 | +// Space: O(h) |
| 3 | + |
| 4 | +/** |
| 5 | + * Definition for a binary tree node. |
| 6 | + * struct TreeNode { |
| 7 | + * int val; |
| 8 | + * TreeNode *left; |
| 9 | + * TreeNode *right; |
| 10 | + * TreeNode(int x) : val(x), left(NULL), right(NULL) {} |
| 11 | + * }; |
| 12 | + */ |
| 13 | + |
| 14 | +class Solution { |
| 15 | +public: |
| 16 | + vector<int> largestValues(TreeNode* root) { |
| 17 | + vector<int> result; |
| 18 | + largestValuesHelper(root, 0, &result); |
| 19 | + return result; |
| 20 | + } |
| 21 | +private: |
| 22 | + void largestValuesHelper(TreeNode* root, int depth, vector<int> *result) { |
| 23 | + if (!root) { |
| 24 | + return; |
| 25 | + } |
| 26 | + if (depth == result->size()) { |
| 27 | + result->emplace_back(root->val); |
| 28 | + } else { |
| 29 | + (*result)[depth] = max((*result)[depth], root->val); |
| 30 | + } |
| 31 | + largestValuesHelper(root->left, depth + 1, result); |
| 32 | + largestValuesHelper(root->right, depth + 1, result); |
| 33 | + } |
| 34 | + }; |
| 35 | + |
| 36 | +// Time: O(n) |
| 37 | +// Space: O(n) |
| 38 | +class Solution2 { |
| 39 | +public: |
| 40 | + vector<int> largestValues(TreeNode* root) { |
| 41 | + if (!root) { |
| 42 | + return {}; |
| 43 | + } |
| 44 | + vector<int> result; |
| 45 | + vector<TreeNode*> curr, next; |
| 46 | + curr.emplace_back(root); |
| 47 | + while (!curr.empty()) { |
| 48 | + int max_val = numeric_limits<int>::min(); |
| 49 | + next.clear(); |
| 50 | + for (const auto& node : curr) { |
| 51 | + max_val = max(max_val, node->val); |
| 52 | + if (node->left) { |
| 53 | + next.emplace_back(node->left); |
| 54 | + } |
| 55 | + if (node->right) { |
| 56 | + next.emplace_back(node->right); |
| 57 | + } |
| 58 | + } |
| 59 | + result.emplace_back(max_val); |
| 60 | + swap(curr, next); |
| 61 | + } |
| 62 | + return result; |
| 63 | + } |
| 64 | +}; |
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