> For the complete documentation index, see [llms.txt](https://protegejj.gitbook.io/algorithm-practice/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://protegejj.gitbook.io/algorithm-practice/leetcode/tree/235-lowest-common-ancestor-of-a-binary-search-tree.md).

# 235 Lowest Common Ancestor of a Binary Search Tree

## 235. Lowest Common Ancestor of a Binary Search Tree

## 1. Question

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](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “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.

## 2. Implementation

**(1) Recursion**

```java
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root.val < Math.min(p.val, q.val)) {
            return lowestCommonAncestor(root.right, p, q);
        }
        else if (root.val > Math.max(p.val, q.val)) {
            return lowestCommonAncestor(root.left, p, q);
        }
        else {
            return root;
        }
    }
}
```

**(2) Iteration**

```java
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        TreeNode curNode = root;
        while (curNode != null) {
            if (curNode.val < p.val && curNode.val < q.val) {
                curNode = curNode.right;
            }
            else if (curNode.val > p.val && curNode.val > q.val) {
                curNode = curNode.left;
            }
            else {
                break;
            }
        }
        return curNode;
    }
}
```

## 3. Time & Space Complexity

Recursion: 时间复杂度: O(h), 空间复杂度: O(h)

Iteration: 时间复杂度:O(h) , 空间复杂度：O(1)