1382_Balance_a_Binary_Search_Tree.cpp

/*
1382. Balance a Binary Search Tree
Given the root of a binary search tree, return a balanced binary search tree with the same node values. If there is more than one answer, return any of them.

A binary search tree is balanced if the depth of the two subtrees of every node never differs by more than 1.

 

Example 1:


Input: root = [1,null,2,null,3,null,4,null,null]
Output: [2,1,3,null,null,null,4]
Explanation: This is not the only correct answer, [3,1,4,null,2] is also correct.
Example 2:


Input: root = [2,1,3]
Output: [2,1,3]
 

Constraints:

The number of nodes in the tree is in the range [1, 104].
1 <= Node.val <= 105
*/
class Solution {
public:
    vector<TreeNode*> sortedArr;
    TreeNode* balanceBST(TreeNode* root) {
        inorderTraverse(root);
        return sortedArrayToBST(0, sortedArr.size() - 1);
    }
    void inorderTraverse(TreeNode* root) {
        if (root == NULL) return;
        inorderTraverse(root->left);
        sortedArr.push_back(root);
        inorderTraverse(root->right);
    }
    TreeNode* sortedArrayToBST(int start, int end) {
        if (start > end) return NULL;
        int mid = (start + end) / 2;
        TreeNode* root = sortedArr[mid];
        root->left = sortedArrayToBST(start, mid - 1);
        root->right = sortedArrayToBST(mid + 1, end);
        return root;
    }
};