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.

思路1.分别和当前节点比较（最开始是根节点）如果p->val和q->val一个大于当前节点值，一个小于当前节点值，则表示当前节点是他们的最低公共祖先，如果两个的大于当前节点的值，则递归调用右子树，如果都小于，则递归调用左子树。

/**

 * 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(p->val > root->val&&q->val > root->val)

        {

           root = lowestCommonAncestor(root->right,p,q);

        }

        if(p->val < root->val&&q->val < root->val)

        {

            root = lowestCommonAncestor(root->left,p,q);

        }

        return root;

    }

};