求一棵普通树的两个结点的最低公共祖先

来源：互联网发布：数据库系统概论编辑：程序博客网时间：2024/05/06 20:55

一棵普通树，树中的结点没有指向父节点的指针，求一棵普通树的两个结点的最低公共祖先。

代码如下，我太懒没有加注释，大家自己看吧！

  1 #include <iostream>  2 #include <list>  3 #include <vector>  4 using namespace std;  5   6 struct TreeNode //节点  7 {  8     char m_nValue;  9     vector<TreeNode*> m_vChildren; 10 }; 11  12 TreeNode* ConstructTree(TreeNode** pNode1 , TreeNode** pNode2) 13 { 14     TreeNode* A = new TreeNode(); 15     A->m_nValue = 'A'; 16     TreeNode* B = new TreeNode(); 17     B->m_nValue = 'B'; 18     TreeNode* C = new TreeNode(); 19     C->m_nValue = 'C'; 20     TreeNode* D = new TreeNode(); 21     D->m_nValue = 'D'; 22     TreeNode* E = new TreeNode(); 23     E->m_nValue = 'E'; 24     TreeNode* F = new TreeNode(); 25     F->m_nValue = 'F'; 26     TreeNode* G = new TreeNode(); 27     G->m_nValue = 'G'; 28     TreeNode* H = new TreeNode(); 29     H->m_nValue = 'H'; 30     TreeNode* I = new TreeNode(); 31     I->m_nValue = 'I'; 32     TreeNode* J = new TreeNode(); 33     J->m_nValue = 'J'; 34  35  36     A->m_vChildren.push_back(B); 37     A->m_vChildren.push_back(C); 38     B->m_vChildren.push_back(D); 39     B->m_vChildren.push_back(E); 40     D->m_vChildren.push_back(F); 41     D->m_vChildren.push_back(G); 42     E->m_vChildren.push_back(H); 43     E->m_vChildren.push_back(I); 44     E->m_vChildren.push_back(J); 45  46     *pNode1 = F; 47     *pNode2 = G; 48     return A; 49 } 50  51  52  53 bool GetNodePath(TreeNode* pRoot , TreeNode* pNode , list<TreeNode*>& path) 54 { 55     if (pRoot == pNode) 56     { 57         return true; 58     } 59     bool find = false ; 60     path.push_back(pRoot); 61     vector<TreeNode*>::iterator it = pRoot->m_vChildren.begin(); 62     while (!find && it != pRoot->m_vChildren.end()) 63     { 64         find = GetNodePath(*it ,pNode , path ) ; 65         it++ ; 66     } 67     if (!find) 68     { 69         path.pop_back(); 70     } 71     return find; 72 } 73  74 TreeNode* GetLastCommonNode(list<TreeNode*>& path1 , list<TreeNode*>& path2 ) 75 { 76     if (path1.empty() || path2.empty()) 77     { 78         return NULL; 79     } 80     list<TreeNode*>::const_iterator it1 = path1.begin(); 81     list<TreeNode*>::const_iterator it2 = path2.begin(); 82     TreeNode* temp = NULL ; 83     while ( it1 != path1.end() && it2 != path2.end() && (*it1 == *it2)) 84     { 85         temp = *it1 ; 86         it1++; 87         it2++; 88     } 89  90     return temp; 91 } 92  93 TreeNode* GetLastCommonParent(TreeNode* pRoot, TreeNode* pNode1, TreeNode* pNode2) 94 { 95     if (pRoot == NULL || pNode1 == NULL || pNode2 == NULL) 96     { 97         return NULL; 98     } 99     list<TreeNode*> path1 ;100     GetNodePath(pRoot, pNode1, path1);101 102     list<TreeNode*> path2 ;103     GetNodePath(pRoot, pNode2, path2);104     105     return GetLastCommonNode(path1, path2);106 }107 108 109 110 int main()111 {112     TreeNode *pNode1 = NULL , *pNode2 = NULL ;113     TreeNode* pRoot = ConstructTree(&pNode1, &pNode2);114     TreeNode* LastCommonParent = GetLastCommonParent(pRoot, pNode1, pNode2);115 116     cout<<LastCommonParent->m_nValue<<endl;117     getchar();118     return 0;119 120 }

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