来自双向链表的平衡二叉搜索树

Question

我正在尝试编写这个函数，它接受一个双链接列表并在适当的位置构造一个平衡的二叉搜索树。 TreeNode.left 相当于前一个指针，TreeNode.right 相当于下一个指针。我从这里的程序中获得灵感，但这不起作用：

http://www.geeksforgeeks.org/in-place-conversion-of-sorted-dll-to-balanced-bst/

private static TreeNode constructBST2(TreeNode head, int m, int n) {
    TreeNode temp = null;
    if (m < n) {
        int mid = m + (n - m)/ 2;
        TreeNode left = constructBST2(head, m, mid);
        temp = head;
        temp.left = left;
        head = head.right;
        temp.right = constructBST2(head, mid + 1, n);
    }
    return temp;
}

Answer 1

让我试试：

private static TreeNode constructBST2(TreeNode root, int r, int m, int n) {
    if (m < n) {
        int leftTreeMid = m + (int)Math.ceil((r - m) / 2);
        int delta = r - leftTreeMid;
        TreeNode left = root;
        for (int i = 0; i < delta; i++)
            left = left.left;
        root.left = left;
        constructBST2(left, leftTreeMid, m, r - 1);

        int rightTreeMid = r + (int)Math.ceil((n - r) / 2);
        delta = rightTreeMid - r;
        TreeNode right = root;
        for(int i = 0; i < delta; i++)
            right = right.right;
        root.right = right;
        constuctBST2(right, rightTreeMid, r + 1, n);
    }
    return root;
}

我根本没试过，也许你可以试试，告诉我它是否有效。