/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
 // Apply divide and conquer and throw the original question to left and right subtree: the subtrees either find the LCA or do not find it. 
 // Definition of the returned value from left and right subtrees:
 // 1. NULL: find no LCA AND not even a node of A or B;
 // 2. A node: either the LCA (both A and B nodes are in this subtree) or one of A and B nodes; to further determine which case it is, we need to look at the returned value of the other subtree: if the other subtree return NULL, then the returned node is LCA (since we are assured A and B nodes are in the tree); if the other subtree return a node, it means A and B nodes locate seperately in both left and right subtrees, and the LCA is the root node.
class Solution {
public:
    /**
     * @param root: The root of the binary search tree.
     * @param A and B: two nodes in a Binary.
     * @return: Return the least common ancestor(LCA) of the two nodes.
     */
    TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *A, TreeNode *B) {
        // write your code here
        // Base case: when root == A (or B) no matter B (or A) locates, the LCA will be the root node;
        if (root == NULL || root == A || root == B) {
            return root;
        }

        TreeNode* left = lowestCommonAncestor(root->left, A, B);
        TreeNode* right = lowestCommonAncestor(root->right, A, B);

        if (left != NULL && right != NULL) {
            return root;
        }

        if (left != NULL && right == NULL) {
            return left;
        }

        if (left == NULL && right != NULL) {
            return right;
        }
    }
};