LeetCode 235 Lowest Common Ancestor of a Binary Search Tree

方法一：坐等左边右边来送结果～～～，适合所有二叉树

 1 class Solution {
 2     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
 3         if (root == null) {
 4             return null;
 5         }
 6         if (root == p || root == q) {
 7             return root;
 8         }
 9 
10         TreeNode left = lowestCommonAncestor(root.left, p, q);
11         TreeNode right = lowestCommonAncestor(root.right, p, q);
12         if (left != null && right != null) {
13             return root;
14         }
15         if (left != null) {
16             return left;
17         }
18         if (right != null) {
19             return right;
20         }
21         return null;
22     }
23 }

方法二 Branch pruning :

转来自老铁的博客https://home.cnblogs.com/u/davidnyc

 1  // the goal is to find the root that would sit in the middle
 2      public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
 3          //base case
 4          if (root == null) return null ;
 5          int min = Math.min(p.val, q.val) ;
 6          int max = Math.max(p.val, q.val) ;
 7          //pre order：看看当前点是不是满足的值
 8          if (root.val>=min && root.val <= max) return root ;
 9          //the left: branch pruning > max, then go left： 左边返回就一层层往上返回
10          if (root.val > max){
11             return lowestCommonAncestor(root.left, p, q) ;
12          }
13          //the right: branch pruning: <min: then go right： 右边返回就一层层往上返回
14          if (root.val < min){
15            return lowestCommonAncestor(root.right, p, q) ;
16          }
17          //不从左边，也不从右边的话，就是NULL
18          return null;
19      }

posted on 2018-03-02 13:51 mayinmiao 阅读(107) 评论(0) 编辑收藏举报