【c++leetcode】1382. Balance a Binary Search Tree

问题入口

DSW (DAY, STOUT & WARREN) ALGORITHM

时间复杂度O(n)

class Solution {
public:
    int makeVine(TreeNode* grand, int cnt = 0)
    {
        auto n = grand->right;
        while (n != nullptr)
        {
            if(n->left != nullptr)
            {
                auto old_n = n;
                n = n->left;
                old_n->left = n->right;
                n->right = old_n;
                grand->right = n;
            }
            else{
                cnt++;
                grand = n;
                n = n->right;
            }
        }
        return cnt;
        
    }
    void compress(TreeNode *grand, int m) {
        auto n = grand->right;
        while (m-- > 0)
        {
            auto old_n = n;
            n = n->right;
            grand->right = n;
            old_n->right = n->left;
            n->left = old_n;
            grand = n;
            n = n->right;
        }
        
    }
    TreeNode* balanceBST(TreeNode *root) {
        TreeNode grand;
        grand.right = root;
        auto cnt = makeVine(&grand);
        int m = pow(2, int(log2(cnt + 1))) - 1;
        compress(&grand, cnt - m);
        for (m = m / 2; m > 0; m /= 2)
            compress(&grand, m);
        return grand.right;
    }
};

1. 将最初的树转变为藤蔓。通过进行右旋转，我们将树展平为“链表”，其中头是以前最左边的节点，尾部是以前最右边的节点。

2. 将树转换为藤蔓后，计算节点总数(cnt)。