strucy TreeNode{TreeNode* left;   //指向左子树TreeNode* right;   //指向右子树TreeNode* father;   //指向父亲节点
};
TreeNode* LowestCommonAncestor(TreeNode* first,TreeNode* second){ }

解析：由于这道题目中树的节点有指向父节点的指针，相对来说是比较好做的。代码如下：

int getHeight(TreeNode *node) {int height = 0;while (node) {height++;node = node->parent;}return height;
}TreeNode* LowestCommonAncestor(TreeNode* first,TreeNode* second) {int height1 = getHeight(first), height2 = getHeight(second), diff = height1 - height2;if (diff < 0) {diff = -diff;while(diff--) {second = second->parent;}} else {while(diff--) {first = first->parent;}}while (first != second) {first = first->parent;second = second->parent;}return first;
}

假设树结构里边并没有指向父节点的指针，那么这个题目应该怎么做呢？剑指offer 上边有关于这种情况的讨论，下边是他山之石：首先获得到两个节点的路径，然后将其转化为寻找路径中最后的公共点，该公共点为两个节点的最低公共祖先。

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;
}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;
}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);
}

今天就到这了，未完待续哦!