二叉树结点的定义如下:
struct BinaryTreeNode { int m_nValue; BinaryTreeNode *m_pLeft; BinaryTreeNode *m_pRight; };
分析:采用递归,先找到值相同的结点,再判断是否相同。
1 #include<stdio.h> 2 3 struct BinaryTreeNode 4 { 5 int m_nValue; 6 BinaryTreeNode* m_pLeft; 7 BinaryTreeNode* m_pRight; 8 }; 9 10 BinaryTreeNode* CreateBinaryTreeNode(int value) 11 { 12 BinaryTreeNode* pNode = new BinaryTreeNode(); 13 pNode->m_nValue = value; 14 pNode->m_pLeft = NULL; 15 pNode->m_pRight = NULL; 16 } 17 18 void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, 19 BinaryTreeNode* pRight) 20 { 21 if(pParent != NULL) 22 { 23 pParent->m_pLeft = pLeft; 24 pParent->m_pRight = pRight; 25 } 26 } 27 28 void DestroyTree(BinaryTreeNode* pRoot) 29 { 30 if(pRoot != NULL) 31 { 32 BinaryTreeNode* pLeft = pRoot->m_pLeft; 33 BinaryTreeNode* pRight = pRoot->m_pRight; 34 35 delete pRoot; 36 pRoot = NULL; 37 38 DestroyTree(pLeft); 39 DestroyTree(pRight); 40 } 41 } 42 43 bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2); 44 45 bool HasSubtree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2) 46 { 47 bool result = false; 48 49 if(pRoot1 != NULL && pRoot2 != NULL) 50 { 51 if(pRoot1->m_nValue == pRoot2->m_nValue) 52 result = DoesTree1HaveTree2(pRoot1, pRoot2); 53 if(!result) 54 result = HasSubtree(pRoot1->m_pLeft, pRoot2); 55 if(!result) 56 result = HasSubtree(pRoot1->m_pRight, pRoot2); 57 } 58 59 return result; 60 } 61 62 bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2) 63 { 64 //如果pRoot2 == NULL 说明树B已经到达叶子结点了 65 //无论pRoot1是否到达一个叶子结点,都应该返回真 66 if(pRoot2 == NULL) 67 return true; 68 69 //如果程序执行到这里,则pRoot2一定不是NULL, 70 //如果pRoot1 == NULL, 则说明树A不包含树B 71 if(pRoot1 == NULL) 72 return false; 73 74 if(pRoot1->m_nValue != pRoot2->m_nValue) 75 return false; 76 77 return DoesTree1HaveTree2(pRoot1->m_pLeft, pRoot2->m_pLeft) && 78 DoesTree1HaveTree2(pRoot1->m_pRight, pRoot2->m_pRight); 79 80 } 81 82 // 树A 树B 83 // 8 8 84 // / \ / 85 // 8 7 9 2 86 // / \ 87 // 9 2 88 // / 89 // 4 7 90 //树B是树A的子树 91 92 int main() 93 { 94 BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8); 95 BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8); 96 BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(7); 97 BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(9); 98 BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(2); 99 BinaryTreeNode* pNodeA6 = CreateBinaryTreeNode(4); 100 BinaryTreeNode* pNodeA7 = CreateBinaryTreeNode(7); 101 102 ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3); 103 ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5); 104 ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7); 105 106 BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8); 107 BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9); 108 BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(2); 109 110 ConnectTreeNodes(pNodeB1, pNodeB2, pNodeB3); 111 112 if(HasSubtree(pNodeA1, pNodeB1)) 113 printf("tree B is Subtree of tree A\n"); 114 else 115 printf("tree B is‘t Subtree of tree A\n"); 116 117 DestroyTree(pNodeA1); 118 DestroyTree(pNodeB1); 119 }
原文:http://www.cnblogs.com/sankexin/p/5616960.html