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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 | +class Solution { |
| 13 | +public: |
| 14 | + void inOrder(TreeNode* root, vector<TreeNode*> &inOrderValues){ |
| 15 | + if(root == NULL) return; |
| 16 | + |
| 17 | + inOrder(root -> left, inOrderValues); |
| 18 | + inOrderValues.push_back(root); |
| 19 | + inOrder(root -> right, inOrderValues); |
| 20 | + } |
| 21 | + |
| 22 | + TreeNode* inorderToBST(int start, int end, vector<TreeNode*> &inOrderValues){ |
| 23 | + if(start > end) return NULL; |
| 24 | + |
| 25 | + int mid = start + (end - start) / 2; |
| 26 | + |
| 27 | + TreeNode* root = inOrderValues[mid]; |
| 28 | + |
| 29 | + root -> left = inorderToBST(start, mid-1, inOrderValues); |
| 30 | + root -> right = inorderToBST(mid + 1, end, inOrderValues); |
| 31 | + |
| 32 | + return root; |
| 33 | + } |
| 34 | + TreeNode* balanceBST(TreeNode* root) { |
| 35 | + vector<TreeNode*> inOrderValues; |
| 36 | + inOrder(root, inOrderValues); |
| 37 | + |
| 38 | + return inorderToBST(0, inOrderValues.size()-1, inOrderValues); |
| 39 | + } |
| 40 | +}; |
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