案例二 树中两个结点的最低公共祖先

      问题: 树中两个结点的最低公共祖先.

    (1)是一颗二叉树，并且是二叉搜素树（根据二叉搜素树的性质求解）

    (2)普通树中结点有指向父结点的指针(演变为两个链表求解第一个公共结点)

    (3)一棵普通的树，树中的结点没有指向父结点的指针(最复杂的情况)

通用的解法如下:

//记录结点的路径bool GetNodePath(TreeNode* pRoot, TreeNode* pNode, list<TreeNode*>& path){    if(pRoot == pNode)        return true;     path.push_back(pRoot);     bool found = false;    vector<TreeNode*>::iterator i = pRoot->m_vChildren.begin();    while(!found && i < pRoot->m_vChildren.end())    {        found = GetNodePath(*i, pNode, path);        ++i;    }     if(!found)        path.pop_back();     return found;}TreeNode* GetLastCommonNode( const list<TreeNode*>& path1,  const list<TreeNode*>& path2){    list<TreeNode*>::const_iterator iterator1 = path1.begin();    list<TreeNode*>::const_iterator iterator2 = path2.begin();        TreeNode* pLast = NULL;     while(iterator1 != path1.end() && iterator2 != path2.end())    {        if(*iterator1 == *iterator2)            pLast = *iterator1;         iterator1++;        iterator2++;    }     return pLast;}//获取最低的公共父结点TreeNode* GetLastCommonParent(TreeNode* pRoot, TreeNode* pNode1, TreeNode* pNode2){    if(pRoot == NULL || pNode1 == NULL || pNode2 == NULL)        return NULL;     list<TreeNode*> path1;    GetNodePath(pRoot, pNode1, path1);     list<TreeNode*> path2;    GetNodePath(pRoot, pNode2, path2);     return GetLastCommonNode(path1, path2);}// ====================测试代码====================void Test(char* testName, TreeNode* pRoot, TreeNode* pNode1, TreeNode* pNode2, TreeNode* pExpected){    if(testName != NULL)        printf("%s begins: \n", testName);    TreeNode* pResult = GetLastCommonParent(pRoot, pNode1, pNode2);    if((pExpected == NULL && pResult == NULL) ||         (pExpected != NULL && pResult != NULL && pResult->m_nValue == pExpected->m_nValue))        printf("Passed.\n");    else        printf("Failed.\n");}// 形状普通的树//              1//            /   \//           2     3//       /       \//      4         5//     / \      / |  \//    6   7    8  9  10void Test1(){    TreeNode* pNode1 = CreateTreeNode(1);    TreeNode* pNode2 = CreateTreeNode(2);    TreeNode* pNode3 = CreateTreeNode(3);    TreeNode* pNode4 = CreateTreeNode(4);    TreeNode* pNode5 = CreateTreeNode(5);    TreeNode* pNode6 = CreateTreeNode(6);    TreeNode* pNode7 = CreateTreeNode(7);    TreeNode* pNode8 = CreateTreeNode(8);    TreeNode* pNode9 = CreateTreeNode(9);    TreeNode* pNode10 = CreateTreeNode(10);    ConnectTreeNodes(pNode1, pNode2);    ConnectTreeNodes(pNode1, pNode3);    ConnectTreeNodes(pNode2, pNode4);    ConnectTreeNodes(pNode2, pNode5);    ConnectTreeNodes(pNode4, pNode6);    ConnectTreeNodes(pNode4, pNode7);    ConnectTreeNodes(pNode5, pNode8);    ConnectTreeNodes(pNode5, pNode9);    ConnectTreeNodes(pNode5, pNode10);    Test("Test1", pNode1, pNode6, pNode8, pNode2);}// 树退化成一个链表//               1//              ///             2//            ///           3//          ///         4//        ///       5void Test2(){    TreeNode* pNode1 = CreateTreeNode(1);    TreeNode* pNode2 = CreateTreeNode(2);    TreeNode* pNode3 = CreateTreeNode(3);    TreeNode* pNode4 = CreateTreeNode(4);    TreeNode* pNode5 = CreateTreeNode(5);    ConnectTreeNodes(pNode1, pNode2);    ConnectTreeNodes(pNode2, pNode3);    ConnectTreeNodes(pNode3, pNode4);    ConnectTreeNodes(pNode4, pNode5);    Test("Test2", pNode1, pNode5, pNode4, pNode3);}// 树退化成一个链表，一个结点不在树中//               1//              ///             2//            ///           3//          ///         4//        ///       5void Test3(){    TreeNode* pNode1 = CreateTreeNode(1);    TreeNode* pNode2 = CreateTreeNode(2);    TreeNode* pNode3 = CreateTreeNode(3);    TreeNode* pNode4 = CreateTreeNode(4);    TreeNode* pNode5 = CreateTreeNode(5);    ConnectTreeNodes(pNode1, pNode2);    ConnectTreeNodes(pNode2, pNode3);    ConnectTreeNodes(pNode3, pNode4);    ConnectTreeNodes(pNode4, pNode5);    TreeNode* pNode6 = CreateTreeNode(6);    Test("Test3", pNode1, pNode5, pNode6, NULL);}// 输入NULLvoid Test4(){    Test("Test4", NULL, NULL, NULL, NULL);}int _tmain(int argc, _TCHAR* argv[]){    Test1();    Test2();    Test3();    Test4();    return 0;}