# How to solve Leetcode 235. Lowest Common Ancestor of a Binary Search Tree

### Problem statement

Given a binary search tree (BST), find the lowest common ancestor (LCA) node 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 `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow a node to be a descendant of itself)."

#### Example 1

![binarysearchtree_improved.png](https://cdn.hashnode.com/res/hashnode/image/upload/v1660338562707/Y-KHp9qcn.png align="left")

```plaintext
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.
```

#### Example 2

![binarysearchtree_improved.png](https://cdn.hashnode.com/res/hashnode/image/upload/v1660338562707/Y-KHp9qcn.png align="left")

```plaintext
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
```

#### Example 3

```plaintext
Input: root = [2,1], p = 2, q = 1
Output: 2
```

#### Constraints

* The number of nodes in the tree is in the range `[2, 10^5]`.
    
* `-10^9 <= Node.val <= 10^9`.
    
* All `Node.val` are unique.
    
* `p != q`.
    
* `p` and `q` will exist in the BST.
    

### Solution: Recursion

Note that in a BST, the values of a `node` and its children `left` and `right` satisfy

```plaintext
left.value < node.value < right.value.
```

It lets you know which branch (left or right) of the `root` the nodes `p` and `q` belong to.

#### Code

```cpp
#include <iostream>
using namespace std;
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if (p->val < root->val && q->val < root->val) {
        return lowestCommonAncestor(root->left, p, q);
    } else if (root->val < p->val && root->val < q->val) {
        return lowestCommonAncestor(root->right, p, q);
    }
    return root;
}

int main() {
    TreeNode zero(0);
    TreeNode three(3);
    TreeNode five(5);
    TreeNode four(4);
    four.left = &three;
    four.right = &five;
    TreeNode two(2);
    two.left = &zero;
    two.right = &four;
    TreeNode seven(7);
    TreeNode nine(9);
    TreeNode eight(8);
    eight.left = &seven;
    eight.right = &nine;
    TreeNode six(6);
    six.left = &two;
    six.right = &eight;

    cout << lowestCommonAncestor(&six, &two, &eight)->val << endl;
    cout << lowestCommonAncestor(&six, &two, &four)->val << endl;
    cout << lowestCommonAncestor(&two, &two, &zero)->val << endl;
}
```

```plaintext
Output:
6
2
2
```

#### Complexity

* Runtime: `O(logN)` (the height of the tree), where `N` is the number of nodes.
    
* Extra space: `O(1)`.
    

### References

* [https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)
    
* [https://www.leetsolve.com/235-lowest-common-ancestor-of-a-binary-search-tree](https://www.leetsolve.com/235-lowest-common-ancestor-of-a-binary-search-tree)