From efbf10e8917eda8993a6f130407c9562b3670dfe Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Sun, 28 Apr 2019 23:53:13 +0800
Subject: [PATCH] Create 109. Convert Sorted List to Binary Search Tree.md

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+# [109. Convert Sorted List to Binary Search Tree](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/)
+# 思路
+根据一个有序链表建立一个平衡的搜索二叉树。        
+## 思路一
+由于需要平衡的，所以每次都应该找到链表中间那个元素作为根节点，找到链表中间节点的方法就是快慢指针法，慢指针每次移动一步，快指针每次移动两步，这样当快指针
+到达链尾时慢指针就处在中间位置。                 
+每次查找需要O(n)的复杂度，次数为进行O(logn)量级，所以总的复杂度应该是O(nlogn)
+
+## 思路二
+为了更快地查找到中间元素，我们还可以考虑将链表元素用一个数组来存放，这样就可以用二分了。           
+每次查找复杂度变成了O(logn)，所以建树总的复杂度应该是O((logn)^2)，但是一开始建立数组时复杂度为O(n)，所以总的时间复杂度应该是O(n)级别（但实测此方法比思路一要慢）
+
+# C++
+## 思路一
+``` C++
+class Solution {
+public:
+    TreeNode *sortedListToBST(ListNode* head) {
+        if (!head) return NULL;
+        if (!head->next) return new TreeNode(head->val);
+        ListNode *slow = head, *fast = head, *last = slow;
+        while (fast->next && fast->next->next) {
+            last = slow;
+            slow = slow->next;
+            fast = fast->next->next;
+        }
+        fast = slow->next;
+        last->next = NULL;
+        TreeNode *cur = new TreeNode(slow->val);
+        if (head != slow) cur->left = sortedListToBST(head);
+        cur->right = sortedListToBST(fast);
+        return cur;
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+private:
+    TreeNode* helper(vector<int>& nodes, int low, int high){
+        if(low > high) return NULL;
+        int mid = (high - low) / 2 + low;
+        TreeNode* root = new TreeNode(nodes[mid]);
+        root -> left = helper(nodes, low, mid - 1);
+        root -> right = helper(nodes, mid + 1, high);
+        return root;
+    }
+public:
+    TreeNode* sortedListToBST(ListNode* head) {
+        vector<int>nodes;
+        while(head){
+            nodes.push_back(head->val);
+            head = head -> next;
+        }
+        return helper(nodes, 0, nodes.size()-1); 
+    }
+};
+```