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| 1 | +/** |
| 2 | + * Definition for a binary tree node. |
| 3 | + * struct TreeNode { |
| 4 | + * int val; |
| 5 | + * TreeNode *left; |
| 6 | + * TreeNode *right; |
| 7 | + * TreeNode() : val(0), left(nullptr), right(nullptr) {} |
| 8 | + * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} |
| 9 | + * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} |
| 10 | + * }; |
| 11 | + */ |
| 12 | + |
| 13 | +class nodeInfo{ |
| 14 | + public: |
| 15 | + int mini; |
| 16 | + int maxi; |
| 17 | + bool isBST; |
| 18 | + int size; |
| 19 | + int sum; |
| 20 | +}; |
| 21 | +class Solution { |
| 22 | +public: |
| 23 | + nodeInfo getLargestBSTSize(TreeNode* root, int &ans){ |
| 24 | + if(root == NULL) return nodeInfo{INT_MAX, INT_MIN, true, 0, 0}; |
| 25 | + |
| 26 | + nodeInfo left = getLargestBSTSize(root -> left, ans); |
| 27 | + nodeInfo right = getLargestBSTSize(root -> right, ans); |
| 28 | + |
| 29 | + nodeInfo currentNode; |
| 30 | + currentNode.size = left.size + right.size + 1; |
| 31 | + currentNode.sum = left.sum + right.sum + root -> val; |
| 32 | + currentNode.mini = min(root -> val, left.mini); |
| 33 | + currentNode.maxi = max(root -> val, right.maxi); |
| 34 | + |
| 35 | + if(left.isBST && right.isBST && (root -> val > left.maxi && root -> val < right.mini)){ |
| 36 | + currentNode.isBST = true; |
| 37 | + ans = max(ans, currentNode.sum); |
| 38 | + }else{ |
| 39 | + currentNode.isBST = false; |
| 40 | + } |
| 41 | + |
| 42 | + return currentNode; |
| 43 | + } |
| 44 | + |
| 45 | + int maxSumBST(TreeNode* root) { |
| 46 | + int max_sum = 0; |
| 47 | + getLargestBSTSize(root, max_sum); |
| 48 | + return max_sum; |
| 49 | + } |
| 50 | +}; |
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