convert sorted array to binary search tree easy

Author: weiy

Tree/ConvertSortedArrayToBinarySearchTree.py

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+"""
+Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
+
+For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
+
+Example:
+
+Given the sorted array: [-10,-3,0,5,9],
+
+One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
+
+      0
+     / \
+   -3   9
+   /   /
+ -10  5
+
+
+ 给定一个已排序过的数组，将它转换为一颗高度平衡的二叉搜索树。
+ 
+ 也就是两颗子树的高度差不超过1。
+
+ 因为是排序的数组，相对来说也异常简单，可以将它看做是一颗二叉搜索树中序遍历后的结果。
+ 按照此结果转换回去就是了。
+
+
+ 每次都二分：
+
+ [-10,-3,0,5,9]
+        mid
+                 5//2 = 2 mid = 2
+            0
+[-10, -3]      [5, 9]   2 // 2 = 1 mid = 1
+    ↓             ↓
+    -3            9
+[-10]  []     [5]    []
+
+为空则不进行下面的操作。
+
+beat 98%
+
+测试地址：
+https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/description/
+
+"""
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution(object):
+    def sortedArrayToBST(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: TreeNode
+        """
+        if not nums:
+            return None
+        
+        def makeBinarySearchTree(nums):
+            mid = len(nums) // 2
+
+            root = TreeNode(nums[mid])
+            left = nums[:mid]
+            right = nums[mid+1:]
+
+            if left:
+                root.left = makeBinarySearchTree(left)
+            if right:
+                root.right = makeBinarySearchTree(right)    
+            
+            return root
+        
+        root = makeBinarySearchTree(nums)
+        return root
+