给定某棵树的预序，有序和后序遍历。任务是检查它们是否都属于同一棵树。
例子：
Input : Inorder -> 4 2 5 1 3
Preorder -> 1 2 4 5 3
Postorder -> 4 5 2 3 1
Output : Yes
Exaplanation : All of the above three traversals are of
the same tree. 1
/ \
2 3
/ \
4 5
Input : Inorder -> 4 2 5 1 3
Preorder -> 1 5 4 2 3
Postorder -> 4 1 2 3 5
Output : No在上一篇文章中，我们已经讨论了通过使用任意两个遍历构造树来解决上述问题的方法。
在本文中，讨论了一种不使用任何额外空间的方法。
方法：
- 在有序数组中搜索前置数组的第一个元素，并将其索引存储为idx，如果不存在则返回False。另外，检查该元素是否存在于后置订单数组中。如果不是，则返回False。
- 对于有序和后置顺序，从第0个索引开始，对于长度为idx的预排序，从第1个索引开始，所有内容都变为预排序数组的第一个元素的左子树。
- 从位置idx + 1到有序和预排序的所有内容，从idx到长度的后序( length-idx-1 )的所有内容，都成为预排序数组的第一个元素的右子树。
- 递归地重复步骤1至3，直到数组长度变为0(在这种情况下，
返回true)或1(在这种情况下，仅当所有三个数组都相等时才返回True，否则返回False)。
下面是上述方法的实现：
C++
// C++ program to check if the given
// three traversals are of the same
// tree or not
#include
using namespace std;
// Function to check if traversals are
// of the same tree
int checktree(int preorder[], int inorder[],
int postorder[], int len)
{
// if the array lengths are 0,
// then all of them are obviously equal
if (len == 0)
return 1;
// if array lengths are 1,
// then check if all of them are equal
if (len == 1)
return (preorder[0] == inorder[0])
&& (inorder[0] == postorder[0]);
// search for first element of preorder
// in inorder array
int idx = -1, f = 0;
for (int i = 0; i < len; ++i)
if (inorder[i] == preorder[0]) {
idx = i;
break;
}
if(idx != -1){
for(int i = 0; i < len; i++)
if(preorder[0] == postorder[i]){f = 1; break;}
}
if (idx == -1 || f == 0)
return 0;
// check for the left subtree
int ret1
= checktree(preorder + 1, inorder, postorder, idx);
// check for the right subtree
int ret2
= checktree(preorder + idx + 1, inorder + idx + 1,
postorder + idx, len - idx - 1);
// return 1 only if both of them are
// correct else 0
return (ret1 && ret2);
}
// Driver Code
int main()
{
// Traversal Arrays
int inorder[] = { 4, 2, 5, 1, 3 };
int preorder[] = { 1, 2, 4, 5, 3 };
int postorder[] = { 4, 5, 2, 3, 1 };
int len1 = sizeof(inorder) / sizeof(inorder[0]);
int len2 = sizeof(preorder) / sizeof(preorder[0]);
int len3 = sizeof(postorder) / sizeof(postorder[0]);
// Check if all the array lengths are equal
if ((len1 == len2) && (len2 == len3)) {
bool res
= checktree(preorder, inorder, postorder, len1);
(res) ? cout << "Yes" : cout << "No";
}
else
cout << "No\n";
return 0;
} Java
// Java program to check if the given
// three traversals are of the same
// tree or not
class GFG {
// Function to check if traversals are
// of the same tree
static boolean checktree(int preorder[], int s,
int inorder[], int s1,
int postorder[], int s2,
int len)
{
// if the array lengths are 0,
// then all of them are obviously equal
if (len == 0)
return true;
// if array lengths are 1,
// then check if all of them are equal
if (len == 1)
return ((preorder[s] == inorder[s1])
&& (inorder[s1] == postorder[s2]));
// search for first element of preorder
// in inorder array
int idx = -1;
for (int i = s1; i < s1 + len; ++i)
if (inorder[i] == preorder[s]) {
idx = i;
break;
}
if (idx == -1)
return false;
idx = idx - s1;
// check for the left subtree
boolean ret1 = checktree(preorder, s + 1, inorder,
s1, postorder, s2, idx);
// check for the right subtree
boolean ret2 = checktree(
preorder, s + idx + 1, inorder, s1 + idx + 1,
postorder, s2 + idx, len - idx - 1);
// return 1 only if both of them are
// correct else 0
return (ret1 && ret2);
}
// Driver Code
public static void main(String args[])
{
// Traversal Arrays
int inorder[] = { 4, 2, 5, 1, 3 };
int preorder[] = { 1, 2, 4, 5, 3 };
int postorder[] = { 4, 5, 2, 3, 1 };
int len1 = inorder.length;
int len2 = preorder.length;
int len3 = postorder.length;
// Check if all the array lengths are equal
if ((len1 == len2) && (len2 == len3)) {
boolean res = checktree(preorder, 0, inorder, 0,
postorder, 0, len1);
System.out.print(((res) ? "Yes" : "No"));
}
else
System.out.print("No\n");
}
}
// This code is contributed by Arnab KunduPython
# Python program to check if the given
# three traversals are of the same
# tree or not
# Function to check if all three traversals
# are of the same tree
def checktree(preorder, inorder, postorder, length):
# if the array lengths are 0,
# then all of them are obviously equal
if length == 0:
return 1
# if array lengths are 1,
# then check if all of them are equal
if length == 1:
return (preorder[0] == inorder[0]) and
(inorder[0] == postorder[0])
# search for first element of preorder
# in inorder array
idx = -1
for i in range(length):
if inorder[i] == preorder[0]:
idx = i
break
if idx == -1:
return 0
# check for the left subtree
ret1 = checktree(preorder[1:], inorder, postorder, idx)
# check for the right subtree
ret2 = checktree(preorder[idx + 1:], inorder[idx + 1:],
postorder[idx:], length-idx-1)
# return 1 only if both of them are correct else 0
return (ret1 and ret2)
# Driver Code
if __name__ == "__main__":
inorder = [4, 2, 5, 1, 3]
preorder = [1, 2, 4, 5, 3]
postorder = [4, 5, 2, 3, 1]
len1 = len(inorder)
len2 = len(preorder)
len3 = len(postorder)
# check if all the array lengths are equal
if (len1 == len2) and (len2 == len3):
correct = checktree(preorder, inorder,
postorder, len1)
if (correct):
print("Yes")
else:
print("No")
else:
print("No")C#
// C# program to check if the given
// three traversals are of the same
// tree or not
using System;
class GFG {
// Function to check if traversals are
// of the same tree
static bool checktree(int[] preorder, int s,
int[] inorder, int s1,
int[] postorder, int s2, int len)
{
// if the array lengths are 0,
// then all of them are obviously equal
if (len == 0)
return true;
// if array lengths are 1,
// then check if all of them are equal
if (len == 1)
return ((preorder[s] == inorder[s1])
&& (inorder[s1] == postorder[s2]));
// search for first element of preorder
// in inorder array
int idx = -1;
for (int i = s1; i < len; ++i)
if (inorder[i] == preorder[s]) {
idx = i;
break;
}
if (idx == -1)
return false;
// check for the left subtree
bool ret1 = checktree(preorder, s + 1, inorder, s1,
postorder, s2, idx);
// check for the right subtree
bool ret2 = checktree(
preorder, s + idx + 1, inorder, s1 + idx + 1,
postorder, s2 + idx, len - idx - 1);
// return 1 only if both of them are
// correct else 0
return (ret1 && ret2);
}
// Driver Code
static public void Main()
{
// Traversal Arrays
int[] inorder = { 4, 2, 5, 1, 3 };
int[] preorder = { 1, 2, 4, 5, 3 };
int[] postorder = { 4, 5, 2, 3, 1 };
int len1 = inorder.Length;
int len2 = preorder.Length;
int len3 = postorder.Length;
// Check if all the array lengths are equal
if ((len1 == len2) && (len2 == len3)) {
bool res = checktree(preorder, 0, inorder, 0,
postorder, 0, len1);
Console.Write(((res) ? "Yes" : "No"));
}
else
Console.Write("No\n");
}
}
// This code is contributed by ajitPHP
输出：
Yes想要从精选的最佳视频中学习和练习问题，请查看《基础知识到高级C的C基础课程》。