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.

//解题思路：
//依据BST的性质，左子树节点的值＜根节点的值。右子树节点的值＞根节点的值
//记当前节点为node，从根节点root出发
//若p与q分别位于node的两側，或当中之中的一个的值与node同样，则node为LCA
//若p的值小于node的值，则LCA位于node的左子树
//否则，LCA位于node的右子树
class Solution {
public:
	TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
		while (root){
			if( (p->val - root->val)*(q->val - root->val)<=0 )  return root;
			else if (p->val < root->val) 
				return lowestCommonAncestor(root->left, p, q);
			else 
				return lowestCommonAncestor(root->right, p, q);
		}
	}
};