面试题50 树中两个结点的最低公共祖先 

 1 bool GetNodePath(TreeNode *pRoot, TreeNode *pNode, list<TreeNode *>&path)
 2 {
 3     if (pRoot == pNode)
 4         return true;
 5     path.push_back(pRoot);
 6     bool found = false;
 7     vector<TreeNode *>::iterator i = pRoot->m_vChildren.begin();
 8     while (!found && i < pRoot->m_vChildren.end())
 9     {
10         found = GetNodePath(*i, pNode, path);
11         ++i;
12     }
13     if (!found)
14         path.pop_back();
15     return found;
16 }
17 
18 TreeNode *GetLastCommonNode(const list<TreeNode*>& path1, const list<TreeNode*>& path2)
19 {
20     list<TreeNode*>::const_iterator iterator1 = path1.begin();
21     list<TreeNode*>::const_iterator iterator2 = path2.begin();
22 
23     TreeNode* pLast = NULL;
24 
25     while (iterator1 != path1.end() && iterator2 != path2.end())
26     {
27         if (*iterator1 == *iterator2)
28             pLast = *iterator1;
29         iterator1++;
30         iterator2++;
31     }
32     return pLast;
33 }
34 
35 TreeNode* GetLastCommonParent(TreeNode* pRoot, TreeNode* pNode1, TreeNode* pNode2)
36 {
37     if (pRoot == NULL || pNode1 == NULL || pNode2 == NULL)
38         return NULL;
39     list<TreeNode*> path1;
40     GetNodePath(pRoot, pNode1, path1);
41 
42     list<TreddNode*> path2;
43     GetNodePath(pRoot, pNode1, pNode2);
44 
45     return GetLastCommonNode(path1, path2);
46 }

 

posted @ 2016-04-10 16:18  早杰  阅读(151)  评论(0)    收藏  举报