树同构问题
编写一个函数来检测两棵树是否同构。如果两棵树中的一棵树可以通过一系列翻转从另一棵树获得，即通过交换多个节点的左子节点和右子节点，则两棵树称为同构树。任何级别的任意数量的节点都可以交换它们的子节点。两棵空树是同构的。
例如，以下两棵树是同构的，并翻转了以下子树：2 和 3、NULL 和 6、7 和 8。
我们同时遍历两棵树。让正在遍历的两棵树的当前内部节点分别为n1和n2 。以 n1 和 n2 为根的子树同构有以下两个条件。
1) n1 和 n2 的数据相同。
2)以下两个之一对于 n1 和 n2 的孩子是正确的
…… a) n1 的左孩子与 n2 的左孩子同构，n1 的右孩子与 n2 的右孩子同构。
…… b) n1 的左孩子与 n2 的右孩子同构，n1 的右孩子与 n2 的左孩子同构。
C++
// A C++ program to check if two given trees are isomorphic
#include
using namespace std;
/* A binary tree node has data, pointer to left and right children */
struct node
{
int data;
struct node* left;
struct node* right;
};
/* Given a binary tree, print its nodes in reverse level order */
bool isIsomorphic(node* n1, node *n2)
{
// Both roots are NULL, trees isomorphic by definition
if (n1 == NULL && n2 == NULL)
return true;
// Exactly one of the n1 and n2 is NULL, trees not isomorphic
if (n1 == NULL || n2 == NULL)
return false;
if (n1->data != n2->data)
return false;
// There are two possible cases for n1 and n2 to be isomorphic
// Case 1: The subtrees rooted at these nodes have NOT been "Flipped".
// Both of these subtrees have to be isomorphic, hence the &&
// Case 2: The subtrees rooted at these nodes have been "Flipped"
return
(isIsomorphic(n1->left,n2->left) && isIsomorphic(n1->right,n2->right))||
(isIsomorphic(n1->left,n2->right) && isIsomorphic(n1->right,n2->left));
}
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
node* newNode(int data)
{
node* temp = new node;
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return (temp);
}
/* Driver program to test above functions*/
int main()
{
// Let us create trees shown in above diagram
struct node *n1 = newNode(1);
n1->left = newNode(2);
n1->right = newNode(3);
n1->left->left = newNode(4);
n1->left->right = newNode(5);
n1->right->left = newNode(6);
n1->left->right->left = newNode(7);
n1->left->right->right = newNode(8);
struct node *n2 = newNode(1);
n2->left = newNode(3);
n2->right = newNode(2);
n2->right->left = newNode(4);
n2->right->right = newNode(5);
n2->left->right = newNode(6);
n2->right->right->left = newNode(8);
n2->right->right->right = newNode(7);
if (isIsomorphic(n1, n2) == true)
cout << "Yes";
else
cout << "No";
return 0;
} Java
// An iterative java program to solve tree isomorphism problem
/* A binary tree node has data, pointer to left and right children */
class Node
{
int data;
Node left, right;
Node(int item)
{
data = item;
left = right;
}
}
class BinaryTree
{
Node root1, root2;
/* Given a binary tree, print its nodes in reverse level order */
boolean isIsomorphic(Node n1, Node n2)
{
// Both roots are NULL, trees isomorphic by definition
if (n1 == null && n2 == null)
return true;
// Exactly one of the n1 and n2 is NULL, trees not isomorphic
if (n1 == null || n2 == null)
return false;
if (n1.data != n2.data)
return false;
// There are two possible cases for n1 and n2 to be isomorphic
// Case 1: The subtrees rooted at these nodes have NOT been
// "Flipped".
// Both of these subtrees have to be isomorphic.
// Case 2: The subtrees rooted at these nodes have been "Flipped"
return (isIsomorphic(n1.left, n2.left) &&
isIsomorphic(n1.right, n2.right))
|| (isIsomorphic(n1.left, n2.right) &&
isIsomorphic(n1.right, n2.left));
}
// Driver program to test above functions
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
// Let us create trees shown in above diagram
tree.root1 = new Node(1);
tree.root1.left = new Node(2);
tree.root1.right = new Node(3);
tree.root1.left.left = new Node(4);
tree.root1.left.right = new Node(5);
tree.root1.right.left = new Node(6);
tree.root1.left.right.left = new Node(7);
tree.root1.left.right.right = new Node(8);
tree.root2 = new Node(1);
tree.root2.left = new Node(3);
tree.root2.right = new Node(2);
tree.root2.right.left = new Node(4);
tree.root2.right.right = new Node(5);
tree.root2.left.right = new Node(6);
tree.root2.right.right.left = new Node(8);
tree.root2.right.right.right = new Node(7);
if (tree.isIsomorphic(tree.root1, tree.root2) == true)
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code has been contributed by Mayank JaiswalPython3
# Python program to check if two given trees are isomorphic
# A Binary tree node
class Node:
# Constructor to create the node of binary tree
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Check if the binary tree is isomorphic or not
def isIsomorphic(n1, n2):
# Both roots are None, trees isomorphic by definition
if n1 is None and n2 is None:
return True
# Exactly one of the n1 and n2 is None, trees are not
# isomorphic
if n1 is None or n2 is None:
return False
if n1.data != n2.data :
return False
# There are two possible cases for n1 and n2 to be isomorphic
# Case 1: The subtrees rooted at these nodes have NOT
# been "Flipped".
# Both of these subtrees have to be isomorphic, hence the &&
# Case 2: The subtrees rooted at these nodes have
# been "Flipped"
return ((isIsomorphic(n1.left, n2.left)and
isIsomorphic(n1.right, n2.right)) or
(isIsomorphic(n1.left, n2.right) and
isIsomorphic(n1.right, n2.left))
)
# Driver program to test above function
n1 = Node(1)
n1.left = Node(2)
n1.right = Node(3)
n1.left.left = Node(4)
n1.left.right = Node(5)
n1.right.left = Node(6)
n1.left.right.left = Node(7)
n1.left.right.right = Node(8)
n2 = Node(1)
n2.left = Node(3)
n2.right = Node(2)
n2.right.left = Node(4)
n2.right.right = Node(5)
n2.left.right = Node(6)
n2.right.right.left = Node(8)
n2.right.right.right = Node(7)
print ("Yes" if (isIsomorphic(n1, n2) == True) else "No")
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)C#
using System;
// An iterative C# program to solve tree isomorphism problem
/* A binary tree node has data, pointer to left and right children */
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right;
}
}
public class BinaryTree
{
public Node root1, root2;
/* Given a binary tree, print its nodes in reverse level order */
public virtual bool isIsomorphic(Node n1, Node n2)
{
// Both roots are NULL, trees isomorphic by definition
if (n1 == null && n2 == null)
{
return true;
}
// Exactly one of the n1 and n2 is NULL, trees not isomorphic
if (n1 == null || n2 == null)
{
return false;
}
if (n1.data != n2.data)
{
return false;
}
// There are two possible cases for n1 and n2 to be isomorphic
// Case 1: The subtrees rooted at these nodes have NOT been
// "Flipped".
// Both of these subtrees have to be isomorphic.
// Case 2: The subtrees rooted at these nodes have been "Flipped"
return (isIsomorphic(n1.left, n2.left)
&& isIsomorphic(n1.right, n2.right))
|| (isIsomorphic(n1.left, n2.right)
&& isIsomorphic(n1.right, n2.left));
}
// Driver program to test above functions
public static void Main(string[] args)
{
BinaryTree tree = new BinaryTree();
// Let us create trees shown in above diagram
tree.root1 = new Node(1);
tree.root1.left = new Node(2);
tree.root1.right = new Node(3);
tree.root1.left.left = new Node(4);
tree.root1.left.right = new Node(5);
tree.root1.right.left = new Node(6);
tree.root1.left.right.left = new Node(7);
tree.root1.left.right.right = new Node(8);
tree.root2 = new Node(1);
tree.root2.left = new Node(3);
tree.root2.right = new Node(2);
tree.root2.right.left = new Node(4);
tree.root2.right.right = new Node(5);
tree.root2.left.right = new Node(6);
tree.root2.right.right.left = new Node(8);
tree.root2.right.right.right = new Node(7);
if (tree.isIsomorphic(tree.root1, tree.root2) == true)
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by Shrikant13Javascript
输出：
Yes时间复杂度：上述解决方案遍历两棵树。所以时间复杂度是 O(min(m,n)*2) 或 O(min(m,n)) 其中 m 和 n 是给定树中的节点数。