For convenience, the rope's length can be substituted for n, because the length is always greater than the number of substring and concatenation nodes in the rope.
为了方便，可以用rope的长度代替n，因为长度总是比rope中的子串和连接节点的个数大。
As I mentioned in the overview section, random character access on a rope with many internal nodes is approximately o (log n), so traversal is o (n log n).
就像我在概述一节中提到过的，在拥有许多内部节点的rope上随机访问字符的时间大约为o (log n)，所以遍历时间为o (nlog n)。
Recall that returning a single character at an arbitrary position within a rope is an o (log n) operation.
返回rope内任意位置字符的操作是个o (log n)操作。
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