递归删除二叉树 [英] recursive delete on a binary tree

问题描述

我试图了解删除二进制搜索树的递归方法如何工作.我在许多地方遇到的代码如下:

I am trying to understand how the recursive method of deletion of a binary search tree works. The code that I came across in many places looks as follows:

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

但是我不明白a)如果例程中没有返回，它将如何工作? b)何时调用free()?我想到了例如一棵树:

I can't understand however a) how does it work if there are no returns in the routine? b) when free() gets to be called? I think about, e.g., such a tree:

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

所以我的理解是我遍历10-> 6-> 5，然后调用destroy_tree(5-> left).因此，if内部的叶为NULL，并且不执行与if相关的内容，因此不会删除5.在这种推理中我会在哪里犯错误?收卷和放卷在这里如何工作?任何帮助，请感激:-)

So my understanding is that I traverse 10->6->5, and then I call destroy_tree(5->left). Therefore, leaf inside if is NULL, and what is if-dependent is not executed, hence 5 is not being deleted. Where do I make mistake in this reasoning? How does winding and unwinding work here? Any help kindly appreciated :-)

推荐答案

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
  }
}

什么都不需要返回...步骤的历史记录"在调用堆栈上，此时看起来像这样:

Nothing is required to return... the "history" of the steps is on the call stack, which looks something like this at that point:

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

因此，leaf_5消失之后，它会返回堆栈并执行destroy_tree(leaf_6->right, which is leaf_8) ...等等...

So after leaf_5 is gone, it goes back up the stack and does destroy_tree(leaf_6->right, which is leaf_8)... etc...

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