摘要： Two elements of a binary search tree (BST) are swapped by mistake.Recover the tree without changing its structure.Note:A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?思路：BST就是对应的InOrder，1) 那么只需要InOrder，维持一个当前的Val，然后看下一个是不是比当前的大，如果不大，则记录位置，直到找到2个位置fi 阅读全文