第一种情况：二叉搜索树
因为是二叉搜索树所以左边一定比右边小，正负号解决问题（这个想法太秀了）

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    while(root != null) {
        if ((p.val - root.val) * (q.val - root.val) <= 0) return root;
        else root = p.val > root.val ? root.right : root.left;
    }
    return root;
}

第二种情况：一般的二叉树

class Solution {
    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);
        if(left == null) return right;//找到一个
        if(right == null) return left;//找到一个
        return root;//找到两个
    }
}
大概思路就是递归