思路
方法一：迭代
1 /**
2 * Definition for a binary tree node.
3 * struct TreeNode {
4 * int val;
5 * TreeNode *left;
6 * TreeNode *right;
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8 * };
9 */
10 class Solution {
11 public:
12 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13 while(root != NULL) {
14 if(p->val < root->val && q->val < root->val) {
15 root = root->left;
16 } else if(p->val > root->val && q->val > root->val) {
17 root = root->right;
18 } else {
19 break;
20 }
21 }
22
23 return root;
24 }
25 };
方法二：递归
1 /**
2 * Definition for a binary tree node.
3 * struct TreeNode {
4 * int val;
5 * TreeNode *left;
6 * TreeNode *right;
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8 * };
9 */
10 class Solution {
11 public:
12 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13 if(root->val < p->val && root->val < q->val)
14 return lowestCommonAncestor(root->right, p, q);
15 if(root->val > p->val && root->val > q->val)
16 return lowestCommonAncestor(root->left, p, q);
17 return root;
18 }
19 };
参考
面试题68 - I. 二叉搜索树的最近公共祖先（迭代 / 递归，清晰图解）