为什么这个算法的空间复杂度是O(1)

Question

大家好:我阅读下面的算法,找到二叉搜索树中两个节点的最低共同祖先.

 /* A binary tree node has data, pointer to left child
   and a pointer to right child */
 struct node
 {
  int data;
  struct node* left;
  struct node* right;
 };

 struct node* newNode(int );

/* Function to find least comman ancestor of n1 and n2 */
int leastCommanAncestor(struct node* root, int n1, int n2)
{
 /* If we have reached a leaf node then LCA doesn't exist
 If root->data is equal to any of the inputs then input is
 not valid. For example 20, 22 in the given figure */
 if(root == NULL || root->data == n1 || root->data == n2)
 return -1;

 /* If any of the input nodes is child of the current node
 we have reached the LCA. For example, in the above figure
 if we want to calculate LCA of 12 and 14, recursion should
 terminate when we reach 8*/
 if((root->right != NULL) &&
  (root->right->data == n1 || root->right->data == n2))
  return root->data;
 if((root->left != NULL) &&
 (root->left->data == n1 || root->left->data == n2))
 return root->data;   

 if(root->data > n1 && root->data < n2)
   return root->data;
 if(root->data > n1 && root->data > n2)
  return leastCommanAncestor(root->left, n1, n2);
 if(root->data < n1 && root->data < n2)
  return leastCommanAncestor(root->right, n1, n2);
}    

注意,上面的函数假定n1小于n2.时间复杂度:O(n)空间复杂度:O(1)

这个算法是递归的,我知道在调用递归函数调用时,函数参数和其他相关寄存器被推送到堆栈,因此需要额外的空间,另一方面,递归深度与大小或高度有关.比如n,树是否更有意义成为O(n)？

感谢您的解释!

Answer 1

该算法涉及尾递归.在您的问题的上下文中,调用者的堆栈帧可以被被调用者重用.换句话说,所有嵌套的函数调用序列都是将结果传递给桶式旅.因此,实际上只需要一个堆栈帧.

有关更多阅读,请参阅Wikipedia上的Tail Call.

Answer 2

尽管您说该算法的递归实现需要 O(n) 空间（因为需要堆栈空间）是正确的，但它仅使用尾递归，这意味着它可以通过循环重新实现以使用 O(1) 空间：

int leastCommanAncestor(struct node* root, int n1, int n2)
    while (1)
    {
     /* If we have reached a leaf node then LCA doesn't exist
     If root->data is equal to any of the inputs then input is
     not valid. For example 20, 22 in the given figure */
     if(root == NULL || root->data == n1 || root->data == n2)
     return -1;

     /* If any of the input nodes is child of the current node
     we have reached the LCA. For example, in the above figure
     if we want to calculate LCA of 12 and 14, recursion should
     terminate when we reach 8*/
     if((root->right != NULL) &&
      (root->right->data == n1 || root->right->data == n2))
      return root->data;
     if((root->left != NULL) &&
     (root->left->data == n1 || root->left->data == n2))
     return root->data;   

     if(root->data > n1 && root->data < n2)
       return root->data;
     if(root->data > n1 && root->data > n2)
      root = root->left;
     else if(root->data < n1 && root->data < n2)
      root = root->right;
    }
}

（请注意，else必须添加以保持语义不变。）