为什么这个算法的空间复杂度是O（1）

大家好：我读下面的算法来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);
}    

请注意，上面的函数假设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，我知道当调用一个递归函数调用时，函数参数和其他相关寄存器被推送到堆栈，所以需要额外的空间，另一方面，递归深度与大小或高度有关那棵树，比如n，是否更有意义是O（n）？

感谢您在这里的任何解释！

---

虽然你说对于算法的递归实现需要O（n）空间是因为所需的堆栈空间，但它只使用尾递归，这意味着它可以重新实现，以便在循环中使用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，必须添加else才能保持语义不变。）

---

该算法涉及尾部递归。 在你的问题的上下文中，调用者的堆栈帧可以被调用者重用。 换句话说，函数调用的所有嵌套顺序都是将结果传递给bucket bucket。 因此，实际上只需要一个堆栈帧。

有关更多阅读，请参阅维基百科的尾巴呼叫。

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