Question

我试图理解删除二叉搜索树的递归方法是如何工作的。我在许多地方遇到的代码如下所示：

void destroy_tree(struct node *leaf)
{
  if( leaf != 0 )
  {
      destroy_tree(leaf->left);
      destroy_tree(leaf->right);
      free( leaf );
  }
}

我无法理解a）如果例程中没有返回，它是如何工作的？ b）当free（）被调用时？我想到，例如，这样一棵树：

                           10
                         /    \
                        6      14
                       / \    /  \
                      5   8  11  18

所以我的理解是我遍历10-> 6-> 5，然后我调用destroy_tree（5->左）。因此，leaf if if为NULL，并且if-dependent不执行，因此5不被删除。我在哪里弄错了这个推理？如何在这里进行卷绕和退绕？任何帮助都很感激： - ）

Answer 1

此时看起来像这样：

void destroy_tree(struct node *leaf_5)
{
  if( leaf_5 != 0 )  // it's not
  {
      destroy_tree(leaf_5->left); // it's NULL so the call does nothing
      destroy_tree(leaf_5->right); // it's NULL so the call does nothing
      free( leaf_5 );  // free here
  }
}

没有必要返回...步骤的“历史”在调用堆栈上，在这一点看起来像这样：

destroy_tree(leaf_10)
  destroy_tree(leaf_10->left, which is leaf_6)
    destroy_tree(leaf_6->left, which is leaf_5)

所以在leaf_5消失之后，它会回到堆栈并执行destroy_tree(leaf_6->right, which is leaf_8) ......等等......

Answer 2

这是该功能基本上如何运作：

void destroy_tree(struct node *leaf)
{
  if( leaf_5 != 0 )  // it's not
  {
      destroy_tree(leaf->left); 
       // Traverse the tree all the way left before any of the code below gets executed.
      destroy_tree(leaf->right); 
       // Traverse the tree all the way right from the final left node before any of 
       //the code below gets executed
      free( leaf );  // Free the final node
  }
}

以下是完整实现递归删除的代码：

void DeleteNode(TreeNode*& tree);
void Delete(TreeNode*& tree, ItemType item);
void TreeType::DeleteItem(ItemType item)
// Calls the recursive function Delete to delete item from tree.
{
Delete(root, item);
}
void Delete(TreeNode*& tree, ItemType item)
// Deletes item from tree.
// Post: item is not in tree.
{
if (item < tree->info)
Delete(tree->left, item); // Look in left subtree.
else if (item > tree->info)
Delete(tree->right, item); // Look in right subtree.
else
DeleteNode(tree); // Node found; call DeleteNode.
}


void GetPredecessor(TreeNode* tree, ItemType& data);
void DeleteNode(TreeNode*& tree)
// Deletes the node pointed to by tree.
// Post: The user's data in the node pointed to by tree is no
// longer in the tree. If tree is a leaf node or has only one
// non-NULL child pointer, the node pointed to by tree is
// deleted; otherwise, the user's data is replaced by its
// logical predecessor and the predecessor's node is deleted.
{
ItemType data;
TreeNode* tempPtr;
tempPtr = tree;
if (tree->left == NULL)
{
tree = tree->right;
delete tempPtr;
}
else if (tree->right == NULL)
{
tree = tree->left;
delete tempPtr;
}
else
{
GetPredecessor(tree->left, data);
tree->info = data;
Delete(tree->left, data); // Delete predecessor node.
}
}

void GetPredecessor(TreeNode* tree, ItemType& data)
// Sets data to the info member of the rightmost node in tree.
{
while (tree->right != NULL)
tree = tree->right;
data = tree->info;
}

二叉树上的递归删除

2 个答案: