The largest element of the binary search tree must be in the right subtree of the root.

That is correct. In a binary search tree, all elements in the left subtree of a node are less than the node itself, and all elements in the right subtree are greater than the node itself. Therefore, the largest element must be in the right subtree of the root, as there cannot be any elements greater than the root in the left subtree.

相关问题

A complete binary search tree is a complete binary tree that satisfies the left-right ordering property of binary search tree as well as the structure of a complete binary tree. Since a complete binary tree can be perfectly stored in an array, we can also store the complete BST in an array. In this lab, you will be given an array of integers as input. The objective is to write a function to_bst(lst) to convert the array to a complete binary tree that stores in an array with the same length. For example, given an array [29, 72, 1, 34, 22], the corresponding complete BST is [34, 22, 72, 1, 29]. Hint: Consider the inorder traversal of the BST. For example: Test Result keys = [29, 72, 1, 34, 22] to_bst(keys) print(keys)

To convert the given array to a complete BST, we need to perform the following steps:

1. Sort the array in ascending order
2. Construct a complete binary tree using the sorted array
3. Perform inorder traversal of the binary tree and store the elements in the original array in the same order as the traversal

Here's the implementation of the to_bst(lst) function in Python:

def to_bst(lst):
    # Sort the input list
    lst.sort()

    # Construct a complete binary tree using the sorted list
    n = len(lst)
    if n == 0:
        return lst
    root = lst[n // 2]
    left_subtree = to_bst(lst[:n // 2])
    right_subtree = to_bst(lst[n // 2 + 1:])
    binary_tree = [root] + left_subtree + right_subtree

    # Perform inorder traversal of the binary tree and store the elements in the original array
    inorder_traversal(binary_tree, lst, 0)

    return lst

def inorder_traversal(binary_tree, lst, i):
    # Perform inorder traversal of the binary tree and store the elements in the original array
    n = len(binary_tree)
    if i >= n:
        return
    inorder_traversal(binary_tree, lst, 2 * i + 1)
    lst[i] = binary_tree[i]
    inorder_traversal(binary_tree, lst, 2 * i + 2)

The to_bst(lst) function takes in the input list and returns the same list after converting it to a complete BST. The function first sorts the input list in ascending order. It then constructs a complete binary tree using the sorted list by recursively dividing the list into two halves and setting the middle element as the root of the binary tree. Finally, the function performs an inorder traversal of the binary tree and stores the elements in the original list in the same order as the traversal.

Assume the postorder of binary tree T is RELATIONDB, the in order of T is REILTABOND,then the preorder of T will be

To determine the preorder of T, we need to first identify the root of the binary tree. In an inorder traversal of a binary tree, the nodes to the left of the root correspond to the left subtree of the root, and the nodes to the right of the root correspond to the right subtree of the root.

From the given inorder sequence REILTABOND, we can see that the root of the binary tree T is "L".

Next, we need to determine the left and right subtrees of the root. Since the postorder traversal of T is RELATIONDB, we know that the last element "B" is the root of the right subtree of "L". To find the root of the left subtree of "L", we need to look for the last element in the postorder sequence that appears before "B" in the inorder sequence.

From the inorder sequence REILTABOND, we can see that the last element before "B" is "T". Therefore, "T" is the root of the left subtree of "L".

Now we can construct the preorder traversal of T by starting at the root "L" and recursively traversing the left and right subtrees in preorder.

The preorder traversal of T is:

- Root: L
- Left subtree: T, R, E, I, L, A
- Right subtree: B, O, N, D

Therefore, the preorder of T is LTRIELABOND.