LeetCode | 0095. Unique Binary Search Trees II不同的二叉搜索树 II【Python】
LeetCode 0095. Unique Binary Search Trees II不同的二叉搜索树 II【Medium】【Python】【分治】【DFS】
Problem
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]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
问题
给定一个整数 n,生成所有由 1 ... n 为节点所组成的二叉搜索树。
示例:
输入: 3
输出:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]
]
解释:
以上的输出对应以下 5 种不同结构的二叉搜索树:
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
思路
分治 递归 DFS
递归构造
从 1-n 中任选一个 i 作为根节点,i 左边的所有节点构成左子树,i 右边的所有节点构成右子树。
时间复杂度:
$$
O(\frac{4n}{n{1/2}})
$$
空间复杂度:
$$
O(\frac{4n}{n{1/2}})
$$
Python3代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def generateTrees(self, n: int) -> List[TreeNode]:
if n == 0:
return []
else:
return self.generateTreesDFS(1, n)
# DFS
def generateTreesDFS(self, left, right):
if left > right:
return [None]
ans = []
for i in range(left, right + 1):
left_nodes = self.generateTreesDFS(left, i - 1) # all possible left subtrees if i is choosen to be a root
right_nodes = self.generateTreesDFS(i + 1, right) # all possible right subtrees if i is choosen to be a root
for left_node in left_nodes:
for right_node in right_nodes:
root = TreeNode(i)
root.left = left_node
root.right = right_node
ans.append(root)
return ans
代码地址
最怕一生碌碌无为,还说平凡难能可贵。