> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/binary_tree_and_divide_conquer/binary-search-tree/lowest-common-ancestor-of-a-binary-search-tree.md).

# 235. Lowest Common Ancestor of a Binary Search Tree

https\://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/

> 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).”

## Thoughts

二叉搜索树上任意两个结点找最底层的共同祖宗结点。利用BST性质当两个结点都小于root时往左树遍历，都大于时往右树遍历，否则当前结点即答案。

## Code

```cpp
/**
 * 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) {
        auto cur = root;
        while (cur != nullptr) {
            if (p->val < cur->val && q->val < cur->val) {
                cur = cur->left;
            } else if (p->val > cur->val && q->val > cur->val) {
                cur = cur->right;
            } else return cur;
        }
        return cur;
    }
};


```

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {    
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        while ((root.val - p.val) * (root.val - q.val) > 0) {
            root = root.val > p.val ? root.left : root.right;
        }

        return root;
    }
}
```

## Analysis

时间复杂度O(lgN).