二叉树插入（按顺序排序）

Question

我已经在互联网上搜索有关此问题的帮助，但我需要帮助。这并不完全是二叉树的普通插入问题，因为我们不能直接处理树结构本身。我的教授自己写了这篇文章，并为我们提供了可以用来编写与二叉树相关的函数的函数。因此，我无法使用节点和指针之类的东西。这也是 C++ 中的。

无论如何，这是我必须编写的递归函数的描述（以及我开始尝试解决该问题）。请注意，它完全返回一棵新树，它实际上并没有向现有树添加某些内容。

tree_t insert_tree(int elt, tree_t tree)
{
    /* 
    // REQUIRES; tree is a sorted binary tree
    // EFFECTS: returns a new tree with elt inserted at a leaf such that 
    //          the resulting tree is also a sorted binary tree.
    //
    //          for example, inserting 1 into the tree:
    //
    //                           4
    //                         /   \
    //                        /      \
    //                       2        5
    //                      / \      / \
    //                         3 
    //                        / \
    //
    // would yield
    //                           4
    //                         /   \
    //                        /      \
    //                       2        5
    //                      / \      / \
    //                     1   3 
    //                    / \ / \
    // 
    // Hint: an in-order traversal of a sorted binary tree is always a
    //       sorted list, and there is only one unique location for
    //       any element to be inserted.
    */

if (elt < elt(tree_left(tree)){
    return insert_tree(tree_left(left));
} else {
    return insert_tree(tree_right(right));
}
}

以下是我们可以使用的函数：

extern bool tree_isEmpty(tree_t tree);
    // EFFECTS: returns true if tree is empty, false otherwise

extern tree_t tree_make();
    // EFFECTS: creates an empty tree.

extern tree_t tree_make(int elt, tree_t left, tree_t right);
    // EFFECTS: creates a new tree, with elt as it's element, left as
    //          its left subtree, and right as its right subtree

extern int tree_elt(tree_t tree);
    // REQUIRES: tree is not empty
    // EFFECTS: returns the element at the top of tree.

extern tree_t tree_left(tree_t tree);
    // REQUIRES: tree is not empty
    // EFFECTS: returns the left subtree of tree

extern tree_t tree_right(tree_t tree);
    // REQUIRES: tree is not empty
    // EFFECTS: returns the right subtree of tree

extern void tree_print(tree_t tree);
    // MODIFIES: cout
    // EFFECTS: prints tree to cout.

Answer 1

插入到零元素树中很容易：

return tree_make(elt, tree_make(), tree_make());

插入到单元素树中也很容易：

tree_t new_node = tree_make(elt, tree_make(), tree_make());
if(elt < tree_elt(tree))
    return tree_make(tree_elt(tree), new_node, tree_right(tree));
else
    return tree_make(tree_elt(tree), tree_left(tree), new_node);

一般来说，要插入新元素，您需要以这种方式重新创建其所有父元素。

---

第 2 部分：递归

我们有基本情况（零元素树）。我们知道如何将新的子树附加到现有树的根上。

那么如何获取新的子树呢？那么，我们将元素插入到当前子树中怎么样？

以下代码始终将新元素附加在树的最左侧，但是一旦您理解了它，纠正起来应该很简单：

tree_t tree_insert(int elt, tree_t tree)
{
    if(tree_empty(tree)) //base case
        return tree_make(elt, tree_make(), tree_make());
    else
        return tree_make( // make a new node
            tree_elt(tree) // with the same value as the current one
            tree_insert(elt, tree_left(tree)) //insert into the left subtree
            tree_right(tree) // keep the right subtree the same
            );
}