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）​​