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酒类网站建设方案,中国工商业联合会,iis中浏览网站是什么,邯郸网站建设公司哪家好树1 树的同构Problem statement: 问题陈述： Write a function to detect if two trees are isomorphic. Two trees are called isomorphic if one of them can be obtained from other by a series of flips, i.e. by swapping left and right children of a numbe…

树1 树的同构

Problem statement:

问题陈述：

Write a function to detect if two trees are isomorphic. Two trees are called isomorphic if one of them can be obtained from other by a series of flips, i.e. by swapping left and right children of a number of nodes. Any number of nodes at any level can have their children swapped.

编写一个函数来检测两棵树是否同构 。如果通过一系列翻转(即通过交换多个节点的左右子节点)可以从另一棵树中获得两棵树，则称为同构树。任何级别的任何数量的节点都可以交换其子级。

Example1:

范例1：

These two trees are isomorphic

这两个树是同构的

Swap left child & right child of 1

交换左孩子和右孩子1

Swap left & right child of 5

交换5个左右孩子

Example 2:

范例2：

Solution:

解：

The conditions which needed to be satisfied are:

需要满足的条件是：

Empty trees are isomorphic
空树是同构的
Roots must be the same
根必须相同
Either left subtree & right subtree of one must be same with the same of other's, or left subtree of one must been same with right subtree of other's & right subtree of one must same with left subtree of other's.
一个的左子树和右子树必须与另一个的左子树相同，或者一个的左子树必须与另一个的右子树相同，并且一个的右子树必须与另一个的左子树相同。

Pre-requisite:

先决条件：

Two Input binary trees (their roots actually), i.e., root1, root2

两个输入二叉树(实际上是其根)，即，root1，root2

FUNCTION isIsomorphic(Node *root1,Node *root2)
1.  Both are empty then it's isomorphic. //condition-1
IF (!root1 && !root2)
return true;
If particularly exactly one of them is empty, 
then they can't be isomorphic.//extension of condition-1
IF(!root1 || !root2)
return false;
2.  If root value are different, they can't be isomorphic//condition-2
IF(root1->data!=root2->data)
return false;
3.  Check condition-3
Recursively checking subtrees
return ( (  isIsomorphic(root1->left,root2->left) && 
isIsomorphic(root1->right,root2->right) ) || 
(isIsomorphic(root1->right,root2->left) &&
isIsomorphic(root1->left,root2->right)));
END FUNCTION

C++ implementation

C ++实现

#include <bits/stdc++.h>
using namespace std;
// TreeNode node type
class TreeNode{
public:             
int val;           //value
TreeNode *left;    //pointer to left child
TreeNode *right;   //pointer to right child
};
// creating new node
TreeNode* newnode(int data)  
{ 
TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode)); 
node->val = data; 
node->left = NULL; 
node->right = NULL; 
return(node); 
}
// function to check isomorphic trees
bool isIsomorphic(TreeNode *root1,TreeNode *root2)
{
if(!root1 && !root2)
return true;
if(!root1 || !root2)
return false;
if(root1->val!=root2->val)
return false;
return ( (isIsomorphic(root1->left,root2->left) && 
isIsomorphic(root1->right,root2->right) )|| 
(isIsomorphic(root1->right,root2->left) && 
isIsomorphic(root1->left,root2->right)));
}
int main(){
cout<<"tree is built as per example-1\n";
TreeNode *root1=newnode(1); 
root1->left= newnode(5); 
root1->right= newnode(2); 
root1->right->right=newnode(3);
root1->left->left=newnode(8); 
root1->left->right=newnode(4);
TreeNode *root2=newnode(1); 
root2->left= newnode(2); 
root2->right= newnode(5); 
root2->right->right=newnode(8);
root2->right->left=newnode(4);
root2->left->right=newnode(3);
if(isIsomorphic(root1,root2))
cout<<"They are isomorphic tree\n";
else
cout<<"They are not isomorphic tree\n";
cout<<"tree is built as per example-2\n";
TreeNode *root3=newnode(1); 
root3->left= newnode(9); 
root3->right= newnode(2); 
root3->right->right=newnode(3);
root3->left->left=newnode(8); 
root3->left->right=newnode(4);
if(isIsomorphic(root1,root3))
cout<<"They are isomorphic tree\n";
else
cout<<"They are not isomorphic tree\n";
return 0;
}

Output

输出量

tree is built as per example-1
They are isomorphic tree
tree is built as per example-2
They are not isomorphic tree

Explanation with example

举例说明

Let's check the example-1

让我们看一下示例1

Nodes are represented by their respective values for better understanding

节点由各自的值表示，以便更好地理解

In the main we call isIsomorphic(root1,root2) //isIsomorphic(1,1)
------------------------------------------------
isIsomorphic(1,1)
root1 is not NULL
root2 is not NULL
root1->data==root2->data
thus it returns
((isIsomorphic(root1->left,root2->left) && isIsomorphic(root1->right,root2->right)) || 
(isIsomorphic(root1->right,root2->left) && isIsomorphic(root1->left,root2->right)));
i.e.,
((isIsomorphic(5,2) && isIsomorphic(2,5)) || 
(isIsomorphic(2,2) && isIsomorphic(5,5)));
Thus call to isIsomorphic(5,2), isIsomorphic(2,5), 
isIsomorphic(2,2), isIsomorphic(5,5)
------------------------------------------------
isIsomorphic(5,2)
root1 is not NULL
root2 is not NULL
root1->data!=root2->data
thus it returns FALSE
------------------------------------------------
isIsomorphic(2,5)
root1 is not NULL
root2 is not NULL
root1->data!=root2->data
thus it returns FALSE
------------------------------------------------
isIsomorphic(2,2)
root1 is not NULL
root2 is not NULL
root1->data==root2->data
thus it returns 
((isIsomorphic(root1->left,root2->left) && isIsomorphic(root1->right,root2->right) ) || 
(isIsomorphic(root1->right,root2->left) && isIsomorphic(root1->left,root2->right)));
i.e.,
((isIsomorphic(NULL,NULL) && isIsomorphic(3,3)) || 
(isIsomorphic(3,NULL) && isIsomorphic(NULL,3)));
Thus call to isIsomorphic(NULL,NULL), isIsomorphic(3,3), 
isIsomorphic(3,NULL), isIsomorphic(NULL,3)
------------------------------------------------
isIsomorphic(NULL,NULL)
root1 is NULL
root2 is NULL
thus it return TRUE
------------------------------------------------
isIsomorphic(3,3)
root1 is not NULL
root2 is not NULL
root1->data==root2->data
thus it returns 
((isIsomorphic(root1->left,root2->left) && isIsomorphic(root1->right,root2->right) ) || 
(isIsomorphic(root1->right,root2->left) && isIsomorphic(root1->left,root2->right)));
i.e.,
((isIsomorphic(NULL,NULL) && (isIsomorphic(NULL,NULL) || 
(isIsomorphic(NULL,NULL) && (isIsomorphic(NULL,NULL)));
Thus call to isIsomorphic(NULL,NULL), (isIsomorphic(NULL,NULL),
(isIsomorphic(NULL,NULL), (isIsomorphic(NULL,NULL))
All (isIsomorphic(NULL,NULL) returns TRUE
Thus, isIsomorphic(3,3) returns TRUE
------------------------------------------------
isIsomorphic(3,NULL)
exactly one root is empty
Thus it returns FALSE
------------------------------------------------
isIsomorphic(NULL,3)
exactly one root is empty
Thus it returns FALSE
------------------------------------------------
Thus ,
isIsomorphic(2,2)
=((isIsomorphic(NULL,NULL) && isIsomorphic(3,3) ) || 
(isIsomorphic(3,NULL) && isIsomorphic(NULL,3)))
=(TRUE && TRUE)|| (FALSE && FALSE)
=TRUE && FLASE
=TRUE
Same way, we can check that isIsomorphic(5,5) returns TRUE
------------------------------------------------
At main:
isIsomorphic(1,1)
=((isIsomorphic(5,2) && isIsomorphic(2,5)) || 
(isIsomorphic(2,2) && isIsomorphic(5,5)))
=((FALSE && FALSE)||(TRUE && TRUE)
=(FALSE && TRUE)
TRUE
Thus these two trees are isomorphic.