# Leetcode Lowest-Common-Ancestor-Of-a-Binary-Search-Tree

#### 2022-07-08 11:53

> ##### Algorithms:
>
> #algorithm #binary_search #DFS
>
> ##### Data structures:
>
> #DS #binary_tree #binary_search_tree
>
> ##### Difficulty:
>
> #coding_problem #difficulty_easy
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> N/A

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)

---

### Problem

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 `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**).”

#### Examples

**Example 1:**

![](https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png)

**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:**

![](https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png)

**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:**

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

#### Constraints

- The number of nodes in the tree is in the range `[2, 105]`.
- `-109 <= Node.val <= 109`
- All `Node.val` are **unique**.
- `p != q`
- `p` and `q` will exist in the BST.

### Thoughts

> [!summary]
> This is a #binary_search

Because of the features of BST, the maximum val of a left sub-tree is smaller than node, so the valid LCA must meet this:

```
root->val >= small && root->val <= big
```

otherwise, search the left or right subtree.

### Solution

```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) {
    // DFS-like recursion
    // Base cases
    int big = max(q->val, p->val);
    int small = min(q->val, p->val);

    if (root->val >= small && root->val <= big) {
      return root;
    } else if (root->val > big) {
      return lowestCommonAncestor(root->left, p, q);
    } else {
      return lowestCommonAncestor(root->right, p, q);
    }
  }
};
```