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 10
void 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
// /
// 5
void 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
// /
// 5
void 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NULL
void Test4()
{
Test("Test4", NULL, NULL, NULL, NULL);
}

int _tmain(int argc, _TCHAR* argv[])
{
Test1();
Test2();
Test3();
Test4();

return 0;
}