二叉搜索树众数：解析与算法实现

2023-12-21 16:07:31

众数，即在一个集合中出现次数最多的元素，在二叉搜索树中寻找众数是一道经典的算法问题，同时也是数据结构与算法课程的必学内容。本文将对二叉搜索树中的众数问题进行深入解析，并提供高效的算法实现。

算法概述

二叉搜索树是一种特殊的二叉树，其中每个节点的值都大于其左子树的所有节点，而小于其右子树的所有节点。这种性质使二叉搜索树非常适合快速查找和插入数据。

在二叉搜索树中寻找众数，本质上是寻找出现次数最多的元素。由于二叉搜索树的性质，我们可以利用其有序性，将问题转化为一个遍历问题。具体算法步骤如下：

中序遍历二叉搜索树，将每个节点的值存储在一个数组中。
对数组进行排序。
遍历排序后的数组，统计每个元素出现的次数。
找出出现次数最多的元素，即众数。

算法实现

def find_mode(root):
    """
    Find the mode(s) of a binary search tree.

    Args:
        root (TreeNode): The root node of the binary search tree.

    Returns:
        list[int]: A list of the mode(s) of the binary search tree.
    """

    # Initialize an empty array to store the values of the nodes in the binary search tree.
    values = []

    # Perform an in-order traversal of the binary search tree and store the values of the nodes in the array.
    def in_order_traversal(root):
        if root is not None:
            in_order_traversal(root.left)
            values.append(root.val)
            in_order_traversal(root.right)

    in_order_traversal(root)

    # Sort the array in ascending order.
    values.sort()

    # Initialize a dictionary to store the frequencies of the values in the array.
    frequencies = {}

    # Count the frequencies of the values in the array.
    for value in values:
        if value not in frequencies:
            frequencies[value] = 0
        frequencies[value] += 1

    # Find the maximum frequency of the values in the array.
    max_frequency = max(frequencies.values())

    # Initialize an empty list to store the mode(s) of the binary search tree.
    modes = []

    # Find the values that have the maximum frequency and add them to the list of modes.
    for value, frequency in frequencies.items():
        if frequency == max_frequency:
            modes.append(value)

    return modes


# Example usage
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.right.left = TreeNode(6)
root.right.right = TreeNode(7)

print(find_mode(root))