# 235&236. Lowest Common Ancestor of a Binary (Search) Tree

**236. Lowest Common Ancestor of a Binary Tree**

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

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:**

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

<pre><code><strong>Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
</strong><strong>Output: 3
</strong><strong>Explanation: The LCA of nodes 5 and 1 is 3.
</strong></code></pre>

**Example 2:**

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

<pre><code><strong>Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
</strong><strong>Output: 5
</strong><strong>Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: root = [1,2], p = 1, q = 2
</strong><strong>Output: 1
</strong></code></pre>

**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 tree.

My Solutions:

```
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    if (root == null || root == p || root == q) return root;
    
    TreeNode left = lowestCommonAncestor(root.left, p, q);
    TreeNode right = lowestCommonAncestor(root.right, p, q);
    
    // it means the p and q has been found on the root's left and right
    if (left != null && right != null) return root;
    if (left != null) return left;
    if (right != null) return right;
    return null;
}
```

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

235和236的区别是235是BST，因此可以只在root的一侧找p和。因为如果p和q的值在root两边，说明root就是结果。

```
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    if (root == null || root == p || root == q) return root;
    if (root.val > p.val && root.val > q.val) {
        return lowestCommonAncestor(root.left, p, q);
    } else if (root.val < p.val && root.val < q.val) {
        return lowestCommonAncestor(root.right, p, q);
    } else {
        return root;
    }
}
```