Leetcode 108. Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

/**
 * 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 helper( nums, 0, nums.size() - 1);
    }
    TreeNode* helper( vector<int>& nums, int lefts, int rights )
    {
        if ( lefts > rights )   
            return nullptr;       
        
        int mid = lefts + ( rights - lefts ) / 2;  // 二分法求取中间值
        TreeNode* root = new TreeNode(nums[mid]); // 将中间值赋值给根节点
        root -> left = helper( nums, lefts, mid -1); // 递归调用,构造左子树
        root -> right = helper( nums, mid + 1, rights);// 递归, 构造右子树
        return root;
    }
};
原文地址:https://www.cnblogs.com/simplepaul/p/6750254.html