Lowest Common Ancestor(LCA)

1.是二叉搜索树

struct TreeNode  
{  
    int val;  
    TreeNode *left,*right;  
};   

TreeNode *LCA(TreeNode *root,TreeNode *p1,TreeNode *p2)  
{  
    while(true)  
    {   
        if(root->val>p1->val&&root->val>p2->val)root=root->left;   
        else if(root->val<p1->val&&root->val<p2->val)root=root->right;  
        else return root;  
    }      
}      

2.不是二叉搜索树，是二叉树
2.1有父指针
分别从两个结点出发，得到两个到根结点的单向链表，转换为求两个单向链表的第一个公共结点。

2.2没有父指针
（1）http://zhedahht.blog.163.com/blog/static/25411174201081263815813/

（2）

TreeNode *LCA(TreeNode *root,TreeNode *p1,TreeNode *p2)  
{  
    if(!root||root==p1||root==p2)return root;  
    TreeNode *left=LCA(root->left,p1,p2);  
    TreeNode *right=LCA(root->right,p1,p2);
    if(left&&right)return root;  
    else if(left)return left;  
    else if(right)return right;  
    else return NULL;  
}  

（3）Tarjan算法

（4）转换为RMQ问题
http://blog.csdn.net/hz5034/article/details/44084087

3.不是二叉树，是树

struct TreeNode
{
    int val;
    vector<TreeNode *>children;
};

bool dfs(TreeNode *root,TreeNode *&p,vector<TreeNode *> &path)
{
    if(!root)return false;
    path.push_back(root);
    if(root==p)return true;
    for(vector<TreeNode *>::iterator i=root->children.begin();i!=root->children.end();++i)
    {
        if(dfs(*i,p,path))return true;
    }
    path.pop_back();
    return false;
}

TreeNode *LCA(TreeNode *root,TreeNode *p1,TreeNode *p2)
{
    if(!root||!p1||!p2)return NULL;
    vector<TreeNode *> path1,path2;
    dfs(root,p1,path1);
    dfs(root,p2,path2);
    TreeNode *res;
    for(int i=0;i<path1.size()&&i<path2.size();++i)
    {
        if(path1[i]==path2[i])res=path1[i];
        else break;
    }
    return res;
}