二进制搜索树迭代前序遍历，无需额外存储

1 回答

• 1

  刚刚制定出预订遍历的解决方案，可能会有效

  Initialize current as root 
  While current is not NULL
  If current does not have right child
    a) print current root
    b) Go to the left, i.e., current = current->left
  Else
    a) print current root
    a) Make the whole right sub-tree of current as the left node of the rightmost child in the left sub tree(inorder predecessor of current)
    b) Go to the left child, i.e., current = current->left
  

  如果算法出现问题，请发表评论

  回复于 2024-04-23T16:41:13+08:00