【Leetcode】108. Convert Sorted Array to Binary Search Tree（左右子树节点平衡）

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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

题目大意：

给出一个有序数组，我们需要计算出一颗高度平衡的BST。每个节点的左右子树上的节点数不超过1。

解题思路：

找出左右子树的中心点mid即可 mid=（r+l+1）/2

其结果答案不唯一。

/**
 * 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 {
private:
    
    TreeNode* dfs(vector<int>& nums, int left, int right){
        if(left == right){
            TreeNode *tmp = new TreeNode(nums[left]);
            return tmp;
        }else if(left>right){
            return nullptr;
        }
        int mid = (right + left + 1)/2;
        TreeNode *root = new TreeNode(nums[mid]);
        
        root->left = dfs(nums, left, mid-1);

        root->right = dfs(nums, mid+1, right);
        
        return root;
        
    }
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return dfs(nums, 0, nums.size()-1);
    }
};