给定两个数组，它们表示用于创建BST的两个键序列。想象一下，我们从每个数组中创建一个二叉搜索树(BST)。我们需要确定两个BST是否相同，而无需实际构建树。
例子：
Let the input arrays be a[] and b[]
Example 1:
a[] = {2, 4, 1, 3} will construct following tree.
2
/ \
1 4
/
3
b[] = {2, 4, 3, 1} will also also construct the same tree.
2
/ \
1 4
/
3
So the output is "True"
Example 2:
a[] = {8, 3, 6, 1, 4, 7, 10, 14, 13}
b[] = {8, 10, 14, 3, 6, 4, 1, 7, 13}
They both construct the same following BST, so output is "True"
8
/ \
3 10
/ \ \
1 6 14
/ \ /
4 7 13 1)比较两个数组的大小。如果不相同，则返回false。
2)比较两个数组的第一个值。如果不相同，则返回false。
3)从每个给定的数组中创建两个列表，以使第一个列表包含的值小于相应数组的第一项。第二个列表包含更大的值。
4)递归检查第一个数组的第一个列表和第二个数组的第一个列表，第二个数组的第一个列表相同。
C++
// C++ program to check if given two arrays represent
// same BST.
#include
using namespace std;
bool sameBSTs(vector aL1,
vector aL2)
{
// Base cases
if (aL1.size() != aL2.size())
return false;
if (aL1.size() == 0)
return true;
if (aL1[0] != aL2[0])
return false;
// Construct two lists from each input array. The first
// list contains values smaller than first value, i.e.,
// left subtree. And second list contains right subtree.
vector aLLeft1 ;
vector aLRight1 ;
vector aLLeft2 ;
vector aLRight2 ;
for (int i = 1; i < aL1.size(); i++)
{
if (aL1[i] < aL1[0])
aLLeft1.push_back(aL1[i]);
else
aLRight1.push_back(aL1[i]);
if (aL2[i] < aL2[0])
aLLeft2.push_back(aL2[i]);
else
aLRight2.push_back(aL2[i]);
}
// Recursively compare left and right
// subtrees.
return sameBSTs(aLLeft1, aLLeft2) &&
sameBSTs(aLRight1, aLRight2);
}
// Driver code
int main()
{
vector aL1;
vector aL2;
aL1.push_back(3);
aL1.push_back(5);
aL1.push_back(4);
aL1.push_back(6);
aL1.push_back(1);
aL1.push_back(0);
aL1.push_back(2);
aL2.push_back(3);
aL2.push_back(1);
aL2.push_back(5);
aL2.push_back(2);
aL2.push_back(4);
aL2.push_back(6);
aL2.push_back(0);
cout << ((sameBSTs(aL1, aL2))?"true":"false")<<"\n";
return 0;
}
// This code is contributed by Arnab Kundu Java
// Java program to check if given two arrays represent
// same BST.
import java.util.ArrayList;
import java.util.Arrays;
public class SameBST {
static boolean sameBSTs(ArrayList aL1,
ArrayList aL2)
{
// Base cases
if (aL1.size() != aL2.size())
return false;
if (aL1.size() == 0)
return true;
if (aL1.get(0) != aL2.get(0))
return false;
// Construct two lists from each input array. The first
// list contains values smaller than first value, i.e.,
// left subtree. And second list contains right subtree.
ArrayList aLLeft1 = new ArrayList();
ArrayList aLRight1 = new ArrayList();
ArrayList aLLeft2 = new ArrayList();
ArrayList aLRight2 = new ArrayList();
for (int i = 1; i < aL1.size(); i++) {
if (aL1.get(i) < aL1.get(0))
aLLeft1.add(aL1.get(i));
else
aLRight1.add(aL1.get(i));
if (aL2.get(i) < aL2.get(0))
aLLeft2.add(aL2.get(i));
else
aLRight2.add(aL2.get(i));
}
// Recursively compare left and right
// subtrees.
return sameBSTs(aLLeft1, aLLeft2) &&
sameBSTs(aLRight1, aLRight2);
}
// Driver code
public static void main(String[] args)
{
ArrayList aL1 =
new ArrayList(Arrays.
asList(3, 5, 4, 6, 1, 0, 2));
ArrayList aL2 =
new ArrayList(Arrays.
asList(3, 1, 5, 2, 4, 6, 0));
System.out.println(sameBSTs(aL1, aL2));
}
} Python3
# Python3 program to check if given two arrays represent
# same BST.
def sameBSTs(aL1, aL2):
# Base cases
if (len(aL1) != len(aL2)):
return False
if (len(aL1) == 0):
return True
if (aL1[0] != aL2[0]):
return False
# Construct two lists from each input array. The first
# list contains values smaller than first value, i.e.,
# left subtree. And second list contains right subtree.
aLLeft1 = []
aLRight1 = []
aLLeft2 = []
aLRight2 = []
for i in range(1, len(aL1)):
if (aL1[i] < aL1[0]):
aLLeft1.append(aL1[i])
else:
aLRight1.append(aL1[i])
if (aL2[i] < aL2[0]):
aLLeft2.append(aL2[i])
else:
aLRight2.append(aL2[i])
# Recursively compare left and right
# subtrees.
return sameBSTs(aLLeft1, aLLeft2) and sameBSTs(aLRight1, aLRight2)
# Driver code
aL1 = []
aL2 = []
aL1.append(3)
aL1.append(5)
aL1.append(4)
aL1.append(6)
aL1.append(1)
aL1.append(0)
aL1.append(2)
aL2.append(3)
aL2.append(1)
aL2.append(5)
aL2.append(2)
aL2.append(4)
aL2.append(6)
aL2.append(0)
if (sameBSTs(aL1, aL2)):
print("true")
else:
print("false")
# This code is contributed by shubhamsingh10C#
// C# program to check if given
// two arrays represent same BST.
using System;
using System.Linq;
using System.Collections.Generic;
public class SameBST
{
static bool sameBSTs(List aL1,
List aL2)
{
// Base cases
if (aL1.Count != aL2.Count)
return false;
if (aL1.Count == 0)
return true;
if (aL1[0] != aL2[0])
return false;
// Construct two lists from each
// input array. The first list contains
// values smaller than first value, i.e.,
// left subtree. And second list contains right subtree.
List aLLeft1 = new List();
List aLRight1 = new List();
List aLLeft2 = new List();
List aLRight2 = new List();
for (int i = 1; i < aL1.Count; i++)
{
if (aL1[i] < aL1[0])
aLLeft1.Add(aL1[i]);
else
aLRight1.Add(aL1[i]);
if (aL2[i] < aL2[0])
aLLeft2.Add(aL2[i]);
else
aLRight2.Add(aL2[i]);
}
// Recursively compare left and right
// subtrees.
return sameBSTs(aLLeft1, aLLeft2) &&
sameBSTs(aLRight1, aLRight2);
}
// Driver code
public static void Main()
{
int []arr1 = {3, 5, 4, 6, 1, 0, 2};
List aL1 = arr1.ToList();
int []arr2 = {3, 1, 5, 2, 4, 6, 0};
List aL2 = arr2.ToList();
Console.WriteLine(sameBSTs(aL1, aL2));
}
}
/* This code contributed by PrinciRaj1992 */ Javascript
输出：
true时间复杂度： O(n * n)
请参考下面的文章O(n)解决方案
在不构建树的情况下检查相同的BST