From e28db21d4a461ac7cca0c79f60dd214f32703370 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Mon, 12 Nov 2018 23:30:52 +0800
Subject: [PATCH] Create 235. Lowest Common Ancestor of a Binary Search Tree.md

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+# [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/)
+# 思路
+求一棵搜索树(BST)中两个节点的最低公共祖先( lowest common ancestor, LCA).  
+求二叉树中节点的最低公共祖先比较麻烦(要用到非递归的后序遍历算法)，但是这里是搜索树，由于搜索树中的节点是排好序的，所以这题就简单了。   
+若有前提p < q, 那么一共有三种情况:
+* 若p > root, 说明LCA应该在root的右子树内;
+* 若q < root, 说明LCA应该在root的左子树内;
+* 否则，即 p < root < q, 那么root就是LCA。(递归出口)
+
+由此可见很容易写出一个递归算法。不过也可以很容易用while循环实现，见下面的代码。
+
+# C++
+``` C++
+class Solution {
+public:
+    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
+        TreeNode *tmp;
+        if(p -> val > q -> val){ // 保证 p -> val <= q -> val
+            tmp = q;
+            q = p;
+            p = tmp;
+        }
+        tmp = root;
+        while(tmp){
+            if(p -> val > tmp -> val) tmp = tmp -> right;
+            else if(q -> val < tmp -> val) tmp = tmp -> left;
+            else return tmp;
+        }
+    }
+};
+```