【LeetCode & 剑指offer刷题】树题17：Convert Sorted Array to Binary Search Tree

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

Convert Sorted Array to Binary Search Tree

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 everynode 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

---

C++

 

/**

 * Definition for a binary tree node.

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

 

/*

方法：分治法（分成处理左右子序列的问题，与快速排序写法类似）

二叉查找树，根结点总是大于左子树所有结点值，小于右子树所有结点值，中序遍历序列为升序排列

 

分析：转化成一个height-balanced的二叉查找树，所以数组中点就是二叉查找树的根结点

写递归思路：先写递归的主体，再写起点输入，再写递归的出口return语句（尤其注意递归的子函数的出口）

*/

class Solution

{

public:

    TreeNode* sortedArrayToBST(vector<int>& nums)

    {

        return buildTree(nums,0,nums.size()-1); //注意索引从0开始

    }

   

    TreeNode* buildTree(vector<int>& nums, int start, int end)

    {

       

        if(start > end) return NULL; //递归子函数的出口（不能取等号，因为单个元素也要分配空间）

       

        int mid = (start + end + 1)/2; //+1是为了向上取整，由于是升序排列，当(start+end)/2不为整数时，取较大的那个数作为根

        TreeNode* root = new TreeNode(nums[mid]);   //构建根结点

        root->left = buildTree(nums, start, mid-1); //构建左子树

        root->right = buildTree(nums, mid+1, end); //构建右子树

       

        return root; //递归原始母函数的出口，返回最顶层的根结点指针

       

    }

};