Merge branch 'master' of github.com:yanglbme/leetcode

Author: chakyam

README.md

@@ -30,6 +30,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 198 | [House Robber](https://github.com/yanglbme/leetcode/tree/master/solution/198.House%20Robber) | `Dynamic Programming` |
 | 203 | [Remove Linked List Elements](https://github.com/yanglbme/leetcode/tree/master/solution/203.Remove%20Linked%20List%20Elements) | `Linked List` |
 | 231 | [Power of Two](https://github.com/yanglbme/leetcode/tree/master/solution/231.Power%20of%20Two) | `Math`, `Bit Manipulation` |
+| 235 | [Lowest Common Ancestor of a Binary Search Tree](https://github.com/yanglbme/leetcode/tree/master/solution/235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree) | `Tree` |
 | 237 | [Delete Node in a Linked List](https://github.com/yanglbme/leetcode/tree/master/solution/237.Delete%20Node%20in%20a%20Linked%20List) | `Linked List` |
 | 344 | [Reverse String](https://github.com/yanglbme/leetcode/tree/master/solution/344.Reverse%20String) | `Two Pointers`, `String` |
 | 876 | [Middle of the Linked List](https://github.com/yanglbme/leetcode/tree/master/solution/876.Middle%20of%20the%20Linked%20List) | `Linked List` |

solution/235.Lowest Common Ancestor of a Binary Search Tree/README.md

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+## 二叉搜索树的最近公共祖先
+### 题目描述
+
+给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。
+
+[百度百科](https://baike.baidu.com/item/%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88/8918834?fr=aladdin)中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（**一个节点也可以是它自己的祖先**）。”
+
+例如，给定如下二叉搜索树:  root = [6,2,8,0,4,7,9,null,null,3,5]
+```
+        _______6______
+       /              \
+    ___2__          ___8__
+   /      \        /      \
+   0      _4       7       9
+         /  \
+         3   5
+```
+
+示例 1:
+```
+输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
+输出: 6 
+解释: 节点 2 和节点 8 的最近公共祖先是 6。
+```
+
+示例 2:
+```
+输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
+输出: 2
+解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。
+```
+
+说明:
+
+- 所有节点的值都是唯一的。
+- p、q 为不同节点且均存在于给定的二叉搜索树中。
+
+### 解法
+如果 root 左子树存在 p，右子树存在 q，那么 root 为最近祖先；
+如果 p、q 均在 root 左子树，递归左子树；右子树同理。
+
+
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        if (root == null || root == p || root == q) {
+            return root;
+        }
+        TreeNode leftNode = lowestCommonAncestor(root.left, p, q);
+        TreeNode rightNode = lowestCommonAncestor(root.right, p, q);
+        if (leftNode != null && rightNode != null) {
+            return root;
+        }
+        return leftNode != null ? leftNode : (rightNode != null) ? rightNode : null;
+        
+    }
+}
+```

solution/235.Lowest Common Ancestor of a Binary Search Tree/Solution.java

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+class Solution {
+    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        if (root == null || root == p || root == q) {
+            return root;
+        }
+        TreeNode leftNode = lowestCommonAncestor(root.left, p, q);
+        TreeNode rightNode = lowestCommonAncestor(root.right, p, q);
+        if (leftNode != null && rightNode != null) {
+            return root;
+        }
+        return leftNode != null ? leftNode : (rightNode != null) ? rightNode : null;
+        
+    }
+}