# 235. Lowest Common Ancestor of a Binary Search Tree

### Leetcode

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

### 題目

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**).”

&#x20;

**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
```

### 解答 <a href="#ti-jie" id="ti-jie"></a>

* **方法一**

Recursive

binary search tree (BST) 指的是左邊子節點的所有值 < 目前節點值 < 右邊子節點的所有值

根據這個特性我們可以判斷 p, q 與 root 的大小決定往樹的左邊或右邊找

```
var lowestCommonAncestor = function(root, p, q) {
    const parentVal = root.val;
    
    const pVal = p.val;
    const qVal = q.val;
    
    // 如果 p, q 都大於根節點，代表要往右邊子樹找
    if (pVal < parentVal && qVal < parentVal) {
        return lowestCommonAncestor(root.left, p, q);
    }
    
    // 如果 p, q 都小於跟節點，代表要往左邊子樹找
    if (pVal > parentVal && qVal > parentVal) {
        return lowestCommonAncestor(root.right, p, q);
    }
    
    // 當 p, q 分站兩邊時，此時的 root 就是我們要找的答案
    return root;
};
```

Runtime: 101 ms, faster than 72.32% of JavaScript online submissions for Lowest Common Ancestor of a Binary Search Tree.

Memory Usage: 52 MB, less than 79.39% of JavaScript online submissions for Lowest Common Ancestor of a Binary Search Tree.

* **方法二**

Iteration

```
var lowestCommonAncestor = function(root, p, q) {
    let node = root;
    const pVal = p.val;
    const qVal = q.val;
    
    while (node) {
        const nodeVal = node.val;
        if (pVal < nodeVal && qVal < nodeVal) {
            node = node.left
        } else if (pVal > nodeVal && qVal > nodeVal) {
            node = node.right
        } else {
            return node;            
        }
    }
};
```

Runtime: 104 ms, faster than 68.91% of JavaScript online submissions for Lowest Common Ancestor of a Binary Search Tree.

Memory Usage: 52.5 MB, less than 30.65% of JavaScript online submissions for Lowest Common Ancestor of a Binary Search Tree.
