题目：

把排序数组转换为高度最小的二叉搜索树

给一个排序数组（从小到大），将其转换为一棵高度最小的排序二叉树。

样例

给出数组 [1,2,3,4,5,6,7], 返回

     4
   /   \
  2     6
 / \    / \
1   3  5   7
挑战

可能有多个答案，返回任意一个即可

解题：

可以看出，这里的数组是所求二叉树，中序遍历的结果，把这个结果还原成树即可。曾经天勤数据结果好像有这一题。

Java程序：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param nums: an integer array
     * @return: a tree node
     */
    public TreeNode sortedArrayToBST(int[] nums) {  
        // write your code here
        if(nums==null)
            return null;
        return buildTree(nums,0,nums.length - 1);
    }
    public TreeNode buildTree(int[] nums,int start,int end){
        if(start>end)
            return null;
        int median = (start+end)/2;
        TreeNode node = new TreeNode(nums[median]);
        node.left = buildTree(nums,start,median-1);
        node.right = buildTree(nums,median+1,end);
    
        return node;
    }
}
View Code

总耗时: 2855 ms

Python程序：

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""
class Solution:
    """
    @param nums: a list of integer
    @return: a tree node
    """
    def sortedArrayToBST(self, nums):
        # write your code here
        if nums==None:
            return None
        return self.buildTree(nums,0,len(nums)-1)
    
    def buildTree(self,nums,start,end):
        if start>end:
            return None
        median = (start+end)//2
        node = TreeNode(nums[median])
        node.left = self.buildTree(nums,start,median-1)
        node.right = self.buildTree(nums,median+1,end)
        return node
View Code

总耗时: 869 ms