Leetcode 95 & 96. Unique Binary Search Trees I & II

最新推荐文章于 2021-09-03 18:46:33 发布

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最新推荐文章于 2021-09-03 18:46:33 发布

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分类专栏： like backtracking leetcode python tree 文章标签： leetcode python

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本文探讨了如何计算不同结构的二叉搜索树的数量，并生成所有可能的二叉搜索树。提供了两种算法实现：一种用于计算给定整数n时，能构建的不同结构的二叉搜索树数量；另一种则生成所有可能的结构。通过递归方法解决了这一问题，为理解二叉树的构造提供了深入的视角。

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Unique Binary Search Trees I

Given n, how many structurally unique BST’s (binary search trees) that store values 1 … n?

Example:

Input: 3
Output: 5

在这里插入图片描述

class Solution(object):
    def numTrees(self, n):
        """
        :type n: int
        :rtype: int
        """
        dp = [0 for i in range(n+1)]
        dp[0] = 1
        dp[1] = 1
        for i in range(2, n+1):
            for j in range(1, i+1):
                dp[i] += dp[j-1] * dp[i-j]
        return dp[n]

** Unique Binary Search Trees II**

Given an integer n, generate all structurally unique BST’s (binary search trees) that store values 1 … n.

Example:

Input: 3
Output:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]
]

Every number in range(1,n) can be a root.
For every root, we can build left trees and right trees by choosing new roots for them.

class Solution:
    def generateTrees(self, n):
        """
        :type n: int
        :rtype: List[TreeNode]
        """
        def generate(start,end):
            if(start>end):
                return [None,]
            alltree=[]
            for i in range(start,end+1):
                lefttrees=generate(start,i-1)
                righttrees=generate(i+1,end)
                for l in lefttrees:
                    for r in righttrees:
                        cur=TreeNode(i)
                        cur.left=l
                        cur.right=r
                        alltree.append(cur)
            return alltree
        return generate(1,n) if n else []