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.

/**
 * 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:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (!root)
			return NULL;
		TreeNode* tmp = root;
		while (tmp)
		{
			if (tmp == p || tmp == q)
			{//假设相等。则返回该结点
				if (tmp == p)
					return p;
				else
					return q;
			}
			else if ((p->val > tmp->val && q->val < tmp->val) || 
				(p->val < tmp->val && q->val > tmp->val))
			{//假设一个大于当前结点。一个小于当前结点。则返回当前结点
				return tmp;
			}
			else if (p->val < tmp->val && q->val < tmp->val)
			{//假设同一时候小于当前结点，则两个结点都在左子树其中
				tmp = tmp->left;
			}
			else if (p->val > tmp->val && q->val > tmp->val)
			{<span style="font-family: Arial, Helvetica, sans-serif;">//假设同一时候大于当前结点，则两个结点都在右子树其中</span>

				tmp = tmp->right;
			}
		}
		return NULL;
    }
};