思路1：

1. 构建从head到node的path，这个递归的思路需要借鉴。

2. 从2条path中得到公共节点，然后就可以利用以前的方法了。

#include 
     
      using namespace std;struct tree{    int m_value;    tree* m_lhs;    tree* m_rhs;};bool GetNodePath(tree* head, tree* node, list
      
       & path){    if (head == node)        return true;    path.push_back(head);    bool found = false;    if (head->m_lhs)        found = GetNodePath(head->m_lhs, node, path);    if (!found && head->m_rhs)        found = GetNodePath(head->m_rhs, node, path);    if (!found)        path.pop_back();    return found;}tree* LastCommonNode(const list
       
        & path1, const list
        
         & path2){    list
         
          ::const_iterator iter1 = path1.begin();    list
          
           ::const_iterator iter2 = path2.begin(); tree* lastnode = nullptr; while (iter1 != path1.end() && iter2 != path2.end()) { if (*iter1 == *iter2) lastnode = *iter1; ++iter1, ++iter2; } return lastnode;}tree* GetCommonParent(tree* head, tree* node1, tree* node2){ if (head == nullptr || node1 == nullptr || node2 == nullptr) return nullptr; list
           
             path1; GetNodePath(head, node1, path1); list
            
              path2; GetNodePath(head, node2, path2); return LastCommonNode(path1, path2);}
            
           
          
         
        
       
      
     

思路2：

也是递归，如果找到node则返回1，找不到则返回0.

TreeNode* lca = NULL;int traverse(TreeNode* p, TreeNode* p1, TreeNode* p2) {    if (!p) return 0;    if (p == p1 || p == p2) return 1;    int res = traverse(p->m_pLeft, p1, p2) + traverse(p->m_pRight, p1, p2);    if (res == 2 && lca == NULL) lca = p;    return res;}

变型：

如果该树为二叉查找树的话，就要稍微变换下思路了。

如果2个节点的值都小于根节点，则在其左边，否则在右边。