0108. 将有序数组转换为二叉搜索树 [Easy] [Convert Sorted Array to Binary Search Tree].md

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.

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

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题目：将有序数组转化为高度平衡的二叉查找树。

思路：利用二分查找的思想，参考解析。将mid值作为根节点，mid左侧的作为左子树元素，mid右侧的作为右子树元素。对左子树右子树分别进行递归二分。

工程代码下载

/**
 * 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) {
        int n = nums.size();
        return helper(nums, 0, n-1);
    }
private:
    TreeNode* helper(vector<int>& nums, int s, int e){
        if(s > e) return nullptr;
        int mid = s + (e - s)/2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = helper(nums, s, mid-1);
        root->right = helper(nums, mid+1, e);
        return root;
    }
};