LeetCode: 108. Convert Sorted Array to Binary Search Tree

题目描述

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: 根据给定的有序数组，生成一种可能的平衡二叉树。

解题思路

中位数作为根节点，中位数左/右边的所有元素作为左/右子树的节点，用同样的方法处理左边和右边的元素，生成左子树和右子树。

AC 代码

/** * 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:
    // 将 nums 数组中的 [beg, end) 区间转换成平衡二叉树 BST.
    TreeNode* sortedArrayToBSTRef(vector<int>& nums, int beg, int end) 
    {
        if(beg >= end) return nullptr;

        // 中位数作为根节点
        int mid = (beg + end)/2;
        TreeNode* root = new TreeNode(nums[mid]);

        // 同样的操作生成左/右子树
        root->left = sortedArrayToBSTRef(nums, beg, mid);
        root->right = sortedArrayToBSTRef(nums, mid+1, end);

        return root;
    }
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) 
    {
        return sortedArrayToBSTRef(nums, 0, nums.size());
    }
};