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Posts tagged as “BST”

花花酱 LeetCode 95. Unique Binary Search Trees II

Problem

https://leetcode.com/problems/unique-binary-search-trees-ii/description/

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

For example,
Given n = 3, your program should return all 5 unique BST’s shown below.

 

Idea: Recursion

for i in 1..n: pick i as root,
left subtrees can be generated in the same way for n_l = 1 … i – 1,
right subtrees can be generated in the same way for n_r = i + 1, …, n
def gen(s, e):
return [tree(i, l, r) for l in gen(s, i – 1) for r in gen(i + 1, e) for i in range(s, e+1)

# of trees:

n = 0: 1
n = 1: 1
n = 2: 2
n = 3: 5
n = 4: 14
n = 5: 42
n = 6: 132

Trees(n) = Trees(0)*Trees(n-1) + Trees(1)*Trees(n-2) + … + Tress(n-1)*Trees(0)

Time complexity: O(3^n)

Space complexity: O(3^n)

C++

Java

Python 3

Solution 2: DP

Java

Related Problems

 

花花酱 LeetCode 783. Minimum Distance Between BST Nodes

Given a Binary Search Tree (BST) with the root node root, return the minimum difference between the values of any two different nodes in the tree.

Example :

Note:

  1. The size of the BST will be between 2 and 100.
  2. The BST is always valid, each node’s value is an integer, and each node’s value is different.

Solution 1: In order traversal 

Time complexity: O(n)

Space complexity: O(n)

C++

Related Problems:

花花酱 LeetCode 315. Count of Smaller Numbers After Self

题目大意:给你一个数组,对于数组中的每个元素,返回一共有多少在它之后的元素比它小。

Problem:

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Return the array [2, 1, 1, 0].

Idea:

Fenwick Tree / Binary Indexed Tree

BST

Solution 1: Binary Indexed Tree (Fenwick Tree)

C++

Java

Solution 2: BST

C++

Java

花花酱 LeetCode 530. Minimum Absolute Difference in BST

Link

Problem:

Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.

Example:

Note: There are at least two nodes in this BST.


Idea:

Sorting via inorder traversal gives us sorted values, compare current one with previous one to reduce space complexity from O(n) to O(h).

Solution:

C++ O(n) space

C++ O(h) space

Java

Python

Related Problems:

  • [解题报告] LeetCode 98. Validate Binary Search Tree

花花酱 LeetCode 98. Validate Binary Search Tree

Problem:

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

  • The left subtree of a node contains only nodes with keys less than the node’s key.
  • The right subtree of a node contains only nodes with keys greater than the node’s key.
  • Both the left and right subtrees must also be binary search trees.

Example 1:

Binary tree [2,1,3], return true.

Example 2:

Binary tree [1,2,3], return false.

Solution 1

Traverse the tree and limit the range of each subtree and check whether root’s value is in the range.

Time complexity: O(n)

Space complexity: O(n)

Note: in order to cover the range of -2^31 ~ 2^31-1, we need to use long or nullable integer.

C++/long

C++/nullable

Java/nullable

Solution 2

Do an in-order traversal, the numbers should be sorted, thus we only need to compare with the previous number.

Time complexity: O(n)

Space complexity: O(n)

C++

Java

Related Problem