LeetCode--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 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:

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

思路：二分法。遞歸找到中點，然後分配左右子樹，直到沒有可分配的就返回空節點。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return sortedArrayToBST(nums,0,nums.size()-1);
    }
    TreeNode* sortedArrayToBST(vector<int>nums,int left,int right){
        if (left>right) return NULL;
        int mid=(left+right)/2;
        TreeNode *cur=new TreeNode(nums[mid]);
        cur->left=sortedArrayToBST(nums,left,mid-1);
        cur->right=sortedArrayToBST(nums,mid+1,right);
        return cur;
    }
};