树的两个结点的最低公共祖先

题目：求一棵普通树上两个结点的最低公共祖先

分析：求出到这两个结点的路径，然后求路径的最后一个公共结点即可。

实现：

bool GetNodePath(TreeNode* pRoot,TreeNode* pNode,list<TreeNode*>&path)  //求结点的路径
{
    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)   //求两个路径的最后公共结点
{
    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;
}

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);
}