《剑指offer》第六十八题：树中两个结点的最低公共祖先

// 面试题68：树中两个结点的最低公共祖先
// 题目：输入两个树结点，求它们的最低公共祖先。

#include <cstdio>
#include "Tree.h"
#include <list>

using namespace std;

// 得到树结点的路径
bool GetNodePath(const TreeNode* pRoot, const TreeNode* pNode, list<const TreeNode*>& path)
{
    if (pRoot == pNode) //找到了树结点
        return true;

    path.push_back(pRoot);

    bool found = false;

    vector<TreeNode*>::const_iterator i = pRoot->m_vChildren.begin();
    while (!found && i < pRoot->m_vChildren.end()) //遍历当前结点的所有子节点
    {
        found = GetNodePath(*i, pNode, path);
        ++i;
    }

    if (!found) //如果不包含找的树结点则删除路径中当前节点
        path.pop_back();

    return found;
}

// 得到最后一个公共节点
const TreeNode* GetLastCommonNode
(
    const list<const TreeNode*>& path1,
    const list<const TreeNode*>& path2
)
{
    list<const TreeNode*>::const_iterator iterator1 = path1.begin();
    list<const TreeNode*>::const_iterator iterator2 = path2.begin();

    const TreeNode* pLast = nullptr;

    while (iterator1 != path1.end() && iterator2 != path2.end())
    {
        if (*iterator1 == *iterator2)
            pLast = *iterator1;

        iterator1++;
        iterator2++;
    }
    return pLast;
}

// 最后一个公共父节点
const TreeNode* GetLastCommonParent(const TreeNode* pRoot, const TreeNode* pNode1, const TreeNode* pNode2)
{
    //第一个节点路径
    list<const TreeNode*> path1;
    GetNodePath(pRoot, pNode1, path1);

    //第二个节点路径
    list<const TreeNode*> path2;
    GetNodePath(pRoot, pNode2, path2);

    return GetLastCommonNode(path1, path2);
}

// ====================测试代码====================
void Test(const char* testName, const TreeNode* pRoot, const TreeNode* pNode1, const TreeNode* pNode2, TreeNode* pExpected)
{
    if (testName != nullptr)
        printf("%s begins: ", testName);

    const TreeNode* pResult = GetLastCommonParent(pRoot, pNode1, pNode2);

    if ((pExpected == nullptr && pResult == nullptr) ||
        (pExpected != nullptr && pResult != nullptr && 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, nullptr);
}

// 输入nullptr
void Test4()
{
    Test("Test4", nullptr, nullptr, nullptr, nullptr);
}

int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();

    return 0;
}

分析：知识迁移能力很重要。