樹1 樹的同構

Problem statement:

問題陳述：

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.

編寫一個函數來檢測兩棵樹是否同構 。 如果通過一系列翻轉(即通過交換多個節點的左右子節點)可以從另一棵樹中獲得兩棵樹，則稱為同構樹。 任何級別的任何數量的節點都可以交換其子級。

Example1:

范例1：

These two trees are isomorphic

這兩個樹是同構的

Swap left child & right child of 1

交換左孩子和右孩子1

Swap left & right child of 5

交換5個左右孩子

Example 2:

范例2：

Solution:

解：

The conditions which needed to be satisfied are:

需要滿足的條件是：

Empty trees are isomorphic

空樹是同構的

Roots must be the same

根必須相同

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.

一個的左子樹和右子樹必須與另一個的左子樹相同，或者一個的左子樹必須與另一個的右子樹相同，并且一個的右子樹必須與另一個的左子樹相同。

Pre-requisite:

先決條件：

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

兩個輸入二叉樹(實際上是其根)，即，root1，root2

FUNCTION isIsomorphic(Node *root1,Node *root2)1. Both are empty then it's isomorphic. //condition-1IF (!root1 && !root2)return true;If particularly exactly one of them is empty, then they can't be isomorphic.//extension of condition-1IF(!root1 || !root2)return false;2. If root value are different, they can't be isomorphic//condition-2IF(root1->data!=root2->data)return false;3. Check condition-3Recursively checking subtreesreturn ( ( isIsomorphic(root1->left,root2->left) && isIsomorphic(root1->right,root2->right) ) || (isIsomorphic(root1->right,root2->left) &&isIsomorphic(root1->left,root2->right))); END FUNCTION .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

C++ implementation

C ++實現

#include <bits/stdc++.h> using namespace std;// TreeNode node type class TreeNode{public: int val; //valueTreeNode *left; //pointer to left childTreeNode *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";elsecout<<"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";elsecout<<"They are not isomorphic tree\n";return 0; }

Output

輸出量

tree is built as per example-1 They are isomorphic tree tree is built as per example-2 They are not isomorphic tree .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

Explanation with example

舉例說明

Let's check the example-1

讓我們看一下示例1

Nodes are represented by their respective values for better understanding

節點由各自的值表示，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 =TRUESame 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) TRUEThus these two trees are isomorphic.

You can check the second example same way & can find returning FALSE

您可以以相同的方式查看第二個示例并找到返回的FALSE

翻譯自: https://www.includehelp.com/icp/check-if-tree-is-isomorphic.aspx

樹1 樹的同構

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