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一刷剑指offer(50)——树中两个结点的最低公共祖先

//GetLastCommonNode用来得到两个路径path1和path2的最后一个公共结点
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;
}


//GetNodePath用来得到从根结点pRoot开始到达结点pNode的路径,这条路径保存在path中
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;
}

//GetLastCommonParent先调用GetNodePath得到pRoot到达pNode1的路径path1,再得到pRoot到达pNode2的路径path2,接着调用GetLastCommonPath得到path1和path2的最后一个公共结点
//将找两个结点最低祖先转换称找两个链表最终公共结点
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);
}