題目連結：
劍指 Offer 68 - II. 二叉樹的最近公共祖先 - 力扣（LeetCode） (leetcode-cn.com)

方法1(易理解，程式碼量大)：

 1 //方法1：用一個HashMap存入所有節點對應的父節點(建立對映關係)，然後去遍歷其中一個節點的所有父節點即
 2     //(包括父節點的父節點等)
 3     
 4     public static TreeNode lowestCommonAncestor1(TreeNode root,TreeNode p,TreeNode q) {
 5         HashMap<TreeNode,TreeNode> fatherMap = new
 
 HashMap<>();
 6         fatherMap.put(root, root);
 7         process1(root,fatherMap);
 8         HashSet<TreeNode> set = new HashSet<>();
 9         set.add(root);
10         TreeNode cur = p;
11         while (cur != fatherMap.get(cur)) {
12             set.add(cur);
13             cur = fatherMap.get(cur);
 
14         }
15         cur = q;
16         while (cur != fatherMap.get(cur)) {
17             if (set.contains(cur) == true) {
18                 return cur;
19             }
20             cur = fatherMap.get(cur);
21         }
22         return cur;
23     }
24     
25     public static void process1(TreeNode root,HashMap<TreeNode,TreeNode> fatherMap) {
 
26         if (root == null) {
27             return;
28         }
29         fatherMap.put(root.left, root);
30         fatherMap.put(root.right, root);
31         process1(root.left,fatherMap);
32         process1(root.right,fatherMap);
33     }

方法2(難理解，程式碼簡潔)：

 1 //方法2：
 2     public static TreeNode lowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q) {
 3         if (root == null || root == p || root == q) {//保證返回值只要null,p,q三者
 4             return root;
 5         }
 6         TreeNode left = lowestCommonAncestor(root.left,p,q);//找其左子樹是否含有p或q
 7         TreeNode right = lowestCommonAncestor(root.right,p,q);//找其右子樹是否含有p或q
 8         if (left != null && right != null) {//如果從root節點的左右子樹都含有p,q,則第一次出現這種情況的節點
 9                                             //就是所要的節點,直接return root;得到結果
10             return root;
11         }
12         return left != null ? left : right;//有p,q優先return p,q;如果都沒有就return null
13     }