LeetCode Lowest Common Ancestor of a Binary Serach Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              
    ___2__          ___8__
   /              /      
   0      _4       7       9
         /  
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

这个题目的意思就是：让你找出两个结点最近的祖先，也就是和他们血缘关系最近的祖先

我的解法：深度优先搜索，再以这个结点，进行搜索，如果以这个结点进行搜索时，发现了我们要求的两个结点，则该结点就是这两个结点最近的 ，不过我的算法的时间复杂度很高。。。。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void dfs(TreeNode *root, TreeNode *p, TreeNode *q,int &i)
    {
        if (root == NULL)return;
        if (root == p || root == q)
            ++i;
        dfs(root->left, p, q,i);
        dfs(root->right, p, q,i);
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        map<TreeNode*,int> visited;
        stack<TreeNode*> sttn;
        TreeNode * curNode;
        sttn.push(root);
        while (!sttn.empty())
        {
            curNode = sttn.top();
            while(curNode->left != NULL&&visited[curNode->left] != 1)
            {
                curNode = curNode->left;
                sttn.push(curNode);
            }
            if (curNode->right != NULL&&visited[curNode->right] != 1)
            {
                curNode = curNode->right;
                sttn.push(curNode);
            }
            else
            {
                int cnt = 0;
                dfs(curNode, p, q,cnt);
                if (cnt == 2)
                    return curNode;
                visited[curNode] = 1;
                sttn.pop();
            }
        }
    }
};