树的子结构判断

最新推荐文章于 2025-08-12 18:32:44 发布

转载最新推荐文章于 2025-08-12 18:32:44 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/wanghui390/p/3772954.html

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#数据结构与算法

题目：输入两棵二叉树A和B，判断B是不是A的子结构。

 1 #include<iostream>
 2 using namespace std;
 3 
 4 struct BinaryTreeNode {
 5     int m_nValue;
 6     BinaryTreeNode* m_pLeft;
 7     BinaryTreeNode* m_pRight;
 8 }
 9 
10 bool DoesTree1HasTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2) {
11     if(pRoot2 == NULL)
12         return true;
13     if(pRoot1 == NULL)
14         return false;
15 
16     if(pRoot2->m_nValue == pRoot1->m_nValue)
17         return false;
18     return DoesTree1HasTree2(pRoot1->m_pLeft, pRoot2->m_pLeft) 
19         && DoesTree1HasTree2(pRoot1->m_pRight, pRoot2->m_pRight);
20 }
21 
22 bool HasSubTree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2) {
23     bool result = false;
24 
25     if(pRoot1 != NULL && pRoot2 != NULL) {
26         if(pRoot2->m_nValue == pRoot1->m_nValue)
27             result = DoesTree1HasTree2(pRoot1, pRoot2);
28         if(!result)
29             result = HasSubTree(pRoot1->m_pLeft, pRoot2);
30         if(!result)
31             result = HasSubTree(pRoot1->m_pRight, pRoot21);
32     }
33 }
34 
35 int main()
36 {
37     
38     return 0;
39 }