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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 | +// dfs solution |
| 15 | +class Solution { |
| 16 | +public: |
| 17 | + int maxLevelSum(TreeNode* root) { |
| 18 | + vector<int> level_sums; |
| 19 | + dfs(root, 0, &level_sums); |
| 20 | + return distance(level_sums.cbegin(), |
| 21 | + max_element(level_sums.cbegin(), level_sums.cend())) + 1; |
| 22 | + } |
| 23 | + |
| 24 | +private: |
| 25 | + void dfs(TreeNode *node, int i, vector<int> *level_sums) { |
| 26 | + if (!node) { |
| 27 | + return; |
| 28 | + } |
| 29 | + if (i == level_sums->size()) { |
| 30 | + level_sums->emplace_back(); |
| 31 | + } |
| 32 | + (*level_sums)[i] += node->val; |
| 33 | + dfs(node->left, i + 1, level_sums); |
| 34 | + dfs(node->right, i + 1, level_sums); |
| 35 | + } |
| 36 | +}; |
| 37 | + |
| 38 | +// Time: O(n) |
| 39 | +// Space: O(w) |
| 40 | +// bfs solution |
| 41 | +class Solution2 { |
| 42 | +public: |
| 43 | + int maxLevelSum(TreeNode* root) { |
| 44 | + int result = 0, level = 1, max_total = numeric_limits<int>::min(); |
| 45 | + queue<TreeNode*> q({root}); |
| 46 | + while (!q.empty()) { |
| 47 | + queue<TreeNode*> next_q; |
| 48 | + int total = 0; |
| 49 | + while (!q.empty()) { |
| 50 | + const auto node = q.front(); q.pop(); |
| 51 | + total += node->val; |
| 52 | + if (node->left) { |
| 53 | + next_q.emplace(node->left); |
| 54 | + } |
| 55 | + if (node->right) { |
| 56 | + next_q.emplace(node->right); |
| 57 | + } |
| 58 | + } |
| 59 | + if (total > max_total) { |
| 60 | + max_total = total; |
| 61 | + result = level; |
| 62 | + } |
| 63 | + q = move(next_q); |
| 64 | + ++level; |
| 65 | + } |
| 66 | + return result; |
| 67 | + } |
| 68 | +}; |
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